Thermodynamics
222 Questions
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th July Morning Shift
$2.4 \mathrm{~g}$ coal is burnt in a bomb calorimeter in excess of oxygen at $298 \mathrm{~K}$ and $1 \mathrm{~atm}$ pressure. The temperature of the calorimeter rises from $298 \mathrm{~K}$ to $300 \mathrm{~K}$. The enthalpy change during the combustion of coal is $-x \mathrm{~kJ} \mathrm{~mol}^{-1}$. The value of $x$ is ___________. (Nearest Integer)
(Given : Heat capacity of bomb calorimeter $20.0 \mathrm{~kJ} \mathrm{~K}^{-1}$. Assume coal to be pure carbon)
Correct Answer: 200
Explanation:
$\mathrm{Q}($ Heat evolved $)=-\frac{\mathrm{C}_{\text {system }} \Delta \mathrm{T}}{\mathrm{n}}$
$\mathrm{n}_{\text {coal }}=\frac{2.4}{12}$
$\mathrm{Q}=\frac{-20(300-298)}{0.2}$
$Q=-200 \mathrm{~kJ} / \mathrm{mol}$
$x=200$
$\mathrm{n}_{\text {coal }}=\frac{2.4}{12}$
$\mathrm{Q}=\frac{-20(300-298)}{0.2}$
$Q=-200 \mathrm{~kJ} / \mathrm{mol}$
$x=200$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th July Evening Shift
While performing a thermodynamics experiment, a student made the following observations.
HCl + NaOH $\to$ NaCl + H2O $\Delta$H = $-$57.3 kJ mol$-$1
CH3COOH + NaOH $\to$ CH3COONa + H2O $\Delta$H = $-$55.3 kJ mol$-$1
The enthalpy of ionization of CH3COOH as calculated by the student is _____________ kJ mol$-$1. (nearest integer)
Correct Answer: 2
Explanation:
(I) $\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$
$ \Delta \mathrm{H}_{1}=-57.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $
(II) $\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$
$ \Delta \mathrm{H}_{2}=-55.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $
Reaction (I) can be written as
$ \text { (III) } \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HCl}+\mathrm{NaOH} $
$ \Delta \mathrm{H}_{3}=57.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $
By adding (II) and (III)
$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaCl} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{HCl} \quad \Delta \mathrm{H}_{\mathrm{r}} \\ &\begin{aligned} \Delta \mathrm{H}_{\mathrm{r}}=\Delta \mathrm{H}_{3}+\Delta \mathrm{H}_{2} &=57.3-55.3 \\ &=2 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \end{aligned} $
$ \Delta \mathrm{H}_{1}=-57.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $
(II) $\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$
$ \Delta \mathrm{H}_{2}=-55.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $
Reaction (I) can be written as
$ \text { (III) } \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HCl}+\mathrm{NaOH} $
$ \Delta \mathrm{H}_{3}=57.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $
By adding (II) and (III)
$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaCl} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{HCl} \quad \Delta \mathrm{H}_{\mathrm{r}} \\ &\begin{aligned} \Delta \mathrm{H}_{\mathrm{r}}=\Delta \mathrm{H}_{3}+\Delta \mathrm{H}_{2} &=57.3-55.3 \\ &=2 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th July Morning Shift
The enthalpy of combustion of propane, graphite and dihydrogen at $298 \mathrm{~K}$ are $-2220.0 \mathrm{~kJ} \mathrm{~mol}^{-1},-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The magnitude of enthalpy of formation of propane $\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)$ is _______________ $\mathrm{kJ} \,\mathrm{mol}^{-1}$. (Nearest integer)
Correct Answer: 104
Explanation:
Enthalpy of combustion of propane, graphite and $\mathrm{H}_{2}$ at $298 \mathrm{~K}$ are
$\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{I}), \Delta \mathrm{H}_{1}=-2220 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{C}($ graphite $)+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}), \quad \Delta \mathrm{H}_{2}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{I}), \quad \Delta \mathrm{H}_{3}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The desired reaction is
$3 \mathrm{C}$ (graphite) + $4 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})$
$\Delta \mathrm{H}_{\mathrm{f}}=3 \Delta \mathrm{H}_{2}+4 \Delta \mathrm{H}_{3}-\Delta \mathrm{H}_{1}$
$ \begin{aligned} &=3(-393.5)+4(-285.8)-(-2220) \\ &=-103.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
$\left|\Delta \mathrm{H}_{\mathrm{f}}\right| \simeq 104 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{I}), \Delta \mathrm{H}_{1}=-2220 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{C}($ graphite $)+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}), \quad \Delta \mathrm{H}_{2}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{I}), \quad \Delta \mathrm{H}_{3}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The desired reaction is
$3 \mathrm{C}$ (graphite) + $4 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})$
$\Delta \mathrm{H}_{\mathrm{f}}=3 \Delta \mathrm{H}_{2}+4 \Delta \mathrm{H}_{3}-\Delta \mathrm{H}_{1}$
$ \begin{aligned} &=3(-393.5)+4(-285.8)-(-2220) \\ &=-103.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
$\left|\Delta \mathrm{H}_{\mathrm{f}}\right| \simeq 104 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 30th June Morning Shift
1.0 mol of monoatomic ideal gas is expanded from state 1 to state 2 as shown in the figure. The magnitude of the work done for the expansion of gas from state 1 to state 2 at 300 K is ____________ J. (Nearest integer)
(Given : R = 8.3 J K$-$1 mol$-$1, ln10 = 2.3, log2 = 0.30)
Correct Answer: 1718
Explanation:
$
\begin{aligned}
&w=-2.303 ~ n R T ~ \log \frac{V_{2}}{V_{1}} \\\\
&P_{1} V_{1}=P_{2} V_{2} \Rightarrow \frac{V_{2}}{V_{1}}=\frac{P_{1}}{P_{2}}=\frac{6}{3}=2 \\\\
&w=-2.303 \times 1 \times 8.3 \times 300 \times \log (2) \\\\
&w=-1718.1 \mathrm{~J}
\end{aligned}
$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 29th June Evening Shift
2.2 g of nitrous oxide (N2O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, $\Delta$U is '$-$x' J. The value of 'x' is ________. [nearest integer]
(Given : atomic mass of N = 14 g mol$-$1 and of O = 16 g mol$-$1. Molar heat capacity of N2O is 100 J K$-$1 mol$-$1)
Correct Answer: 195
Explanation:
$\Delta T=-40 \mathrm{~K}$
$ \begin{aligned} \Delta U &=q+w \\\\ &=\frac{100 \times 2.2}{44}(-40)-(-49.39) \times 10^{-3} \times 101.325 \end{aligned} $
$ \begin{aligned} &=-200+5 \\\\ &=-195 \mathrm{~J} \\\\ \mathrm{x}=& 195 \end{aligned} $
$ \begin{aligned} \Delta U &=q+w \\\\ &=\frac{100 \times 2.2}{44}(-40)-(-49.39) \times 10^{-3} \times 101.325 \end{aligned} $
$ \begin{aligned} &=-200+5 \\\\ &=-195 \mathrm{~J} \\\\ \mathrm{x}=& 195 \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 29th June Morning Shift
17.0 g of NH3 completely vapourises at $-$33.42$^\circ$C and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol$-$1. The enthalpy change for the vapourisation of 85 g of NH3 under the same conditions is _________ kJ.
Correct Answer: 117
Explanation:
Number of moles of $\mathrm{NH}_{3}=5$
So, required $\Delta \mathrm{H}=5 \times 23.4$
$ =117 \mathrm{~kJ} $
So, required $\Delta \mathrm{H}=5 \times 23.4$
$ =117 \mathrm{~kJ} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th June Evening Shift
For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, $\Delta$CH$\Theta $ = $-$601.70 kJ mol$-$1, the magnitude of change in internal energy for the reaction is __________ kJ. (Nearest integer)
(Given : R = 8.3 J K$-$1 mol$-$1)
Correct Answer: 600
Explanation:
$\mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{MgO}(\mathrm{s})$
$ \begin{aligned} &\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{ngRT} \\\\ &\Delta \mathrm{ng}=-\frac{1}{2} \\\\ &-601.70=\Delta \mathrm{U}-\frac{1}{2}(8.3)(300) \times 10^{-3} \\\\ &\Delta \mathrm{U}=-601.70+1.245 \\\\ &\Delta \mathrm{U} \simeq-600 \mathrm{~kJ} \end{aligned} $
The magnitude of change in internal energy is $600 \mathrm{~kJ}$.
$ \begin{aligned} &\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{ngRT} \\\\ &\Delta \mathrm{ng}=-\frac{1}{2} \\\\ &-601.70=\Delta \mathrm{U}-\frac{1}{2}(8.3)(300) \times 10^{-3} \\\\ &\Delta \mathrm{U}=-601.70+1.245 \\\\ &\Delta \mathrm{U} \simeq-600 \mathrm{~kJ} \end{aligned} $
The magnitude of change in internal energy is $600 \mathrm{~kJ}$.
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th June Morning Shift
4.0 L of an ideal gas is allowed to expand isothermally into vacuum until the total volume is 2.0 L. The amount of heat absorbed in this expansion is ____________ L atm.
Correct Answer: 0
Explanation:
Work done $=-P_{\text {ext }} \Delta \mathrm{v}$
$ \because P_{\text {ext }}=0 \quad \text { (vacuum) } $
$\therefore w=0, \quad\Delta U=0$ (as the process is isothermal)
So, $q=0$
$ \because P_{\text {ext }}=0 \quad \text { (vacuum) } $
$\therefore w=0, \quad\Delta U=0$ (as the process is isothermal)
So, $q=0$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Evening Shift
When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is __________ J. [nearest integer] (Given : R = 8.3 J K$-$1 mol$-$1 and log 2 = 0.3010)
Correct Answer: 8630
Explanation:
$
\begin{aligned}
\mathrm{W}_{\mathrm{rev}} &=-2.303 ~\mathrm{nRT} ~\log _{10}\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right) \\\\
&=-2.303 \times 5 \times 8.3 \times 300 \times \log _{10}\left(\frac{20}{10}\right) \\\\
& \simeq-8630 \mathrm{~J}
\end{aligned}
$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Evening Shift
A fish swimming in water body when taken out from the water body is covered with a film of water of weight 36 g. When it is subjected to cooking at 100$^\circ$C, then the internal energy for vaporization in kJ mol$-$1 is ___________. [nearest integer]
[Assume steam to be an ideal gas. Given $\Delta$vapH$^\Theta $ for water at 373 K and 1 bar is 41.1 kJ mol$-$1 ; R = 8.31 J K$-$1 mol$-$1]
Correct Answer: 38
Explanation:
$\underset{36 \mathrm{~g}}{\mathrm{H}_{2} \mathrm{O}}(\ell) \longrightarrow \underset{36 \mathrm{~g}}{\mathrm{H}_{2} \mathrm{O}}(\mathrm{g})$ (evaporation)
$ \begin{aligned} \mathrm{n}_{\mathrm{H}_{2} \mathrm{O}} &=\frac{36}{18}=2 \quad \Delta \mathrm{n}_{\mathrm{g}}=1-0=1 \\\\ \Delta \mathrm{U}_{\text {vap }} &=\Delta \mathrm{H}_{\text {vap }}-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\\\ &=41.1-(1) \times 8.31 \times 10^{-3} \times 373 \\\\ &=41.1-3.099 \\\\ &=38 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $
$ \begin{aligned} \mathrm{n}_{\mathrm{H}_{2} \mathrm{O}} &=\frac{36}{18}=2 \quad \Delta \mathrm{n}_{\mathrm{g}}=1-0=1 \\\\ \Delta \mathrm{U}_{\text {vap }} &=\Delta \mathrm{H}_{\text {vap }}-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\\\ &=41.1-(1) \times 8.31 \times 10^{-3} \times 373 \\\\ &=41.1-3.099 \\\\ &=38 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Morning Shift
For complete combustion of methanol
CH3OH(I) + ${3 \over 2}$O2(g) $\to$ CO2(g) + 2H2O(I)
the amount of heat produced as measured by bomb calorimeter is 726 kJ mol$-$1 at 27$^\circ$C. The enthalpy of combustion for the reaction is $-$x kJ mol$-$1, where x is ___________. (Nearest integer)
(Given : R = 8.3 JK$-$1 mol$-$1)
Correct Answer: 727
Explanation:
$\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
$ \begin{aligned} &\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\\\ &=-726 \mathrm{~kJ}+\left(\frac{-1}{2}\right) \times 8.3 \times 300 \\\\ &\simeq-727 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
$ \begin{aligned} &\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\\\ &=-726 \mathrm{~kJ}+\left(\frac{-1}{2}\right) \times 8.3 \times 300 \\\\ &\simeq-727 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Morning Shift
The standard entropy change for the reaction
4Fe(s) + 3O2(g) $\to$ 2Fe2O3(s) is $-$550 J K$-$1 at 298 K.
[Given : The standard enthalpy change for the reaction is $-$165 kJ mol$-$1]. The temperature in K at which the reaction attains equilibrium is _____________. (Nearest Integer)
Correct Answer: 300
Explanation:
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=0$ at equilibrium
$ \begin{aligned} &\Rightarrow-165 \times 10^3-\mathrm{T} \times(-505)=0 \\\\ &\Rightarrow \mathrm{T}=300 \mathrm{~K} \end{aligned} $
$ \begin{aligned} &\Rightarrow-165 \times 10^3-\mathrm{T} \times(-505)=0 \\\\ &\Rightarrow \mathrm{T}=300 \mathrm{~K} \end{aligned} $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Evening Shift
The incorrect expression among the following is :
A.
${{\Delta {G_{System}}} \over {\Delta {S_{Total}}}} = - T$ (at constant P)
B.
$K = {{\Delta H^\circ - T\Delta S^\circ } \over {RT}}$
C.
$K = {e^{ - \Delta G^\circ /RT}}$
D.
For isothermal process ${w_{reversible}} = - nRT\ln {{{V_f}} \over {{V_i}}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
During which of the following processes, does entropy decrease?
(A) Freezing of water to ice at 0$^\circ$C
(B) Freezing of water to ice at $-$10$^\circ$C
(C) N2(g) + 3H2(g) $ \to $ 2NH3(g)
(D) Adsorption of CO(g) on lead surface.
(E) Dissolution of NaCl in water
Choose the correct answer from the options given below :
(A) Freezing of water to ice at 0$^\circ$C
(B) Freezing of water to ice at $-$10$^\circ$C
(C) N2(g) + 3H2(g) $ \to $ 2NH3(g)
(D) Adsorption of CO(g) on lead surface.
(E) Dissolution of NaCl in water
Choose the correct answer from the options given below :
A.
(A), (C) and (E) only
B.
(B) and (C) only
C.
(A), (B), (C) and (D) only
D.
(A) and (E) only
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
For the reaction, 2NO2(g) $\rightleftharpoons$ N2O4(g), when $\Delta$S = $-$176.0 JK$-$1 and $\Delta$H = $-$57.8 kJ mol$-$1, the magnitude of $\Delta$G at 298 K for the reaction is ___________ kJ mol$-$1. (Nearest integer)
Correct Answer: 5
Explanation:
Given, $\Delta$H = $-$ 57.8 kJ mol$-$1
$\Delta$S = $-$ 176 JK$-$1 mol$-$1
T = 298 K
Using Gibb's free energy relation
$\Delta$G = $\Delta$H $-$ T$\Delta$S
where, $\Delta$G = change in Gibb's free energy
$\Delta$H = change in enthalpy
T = temperature
$\Delta$S = change in entropy
$\Delta$G = 57.8 kJ/mol $-$ [298 K $\times$ ($-$ 176 Jk$-$1 mol$-$1)]
= 57.8 kJ/mol $-$ $\left( {298 \times {{ - 176} \over {1000}}kJ} \right)$ [$\therefore$ 1 kJ = 1000 J]
= $-$ 5.352 kJ/mol
| $\Delta$G | = 5.352
Hence, answer is 5.
$\Delta$S = $-$ 176 JK$-$1 mol$-$1
T = 298 K
Using Gibb's free energy relation
$\Delta$G = $\Delta$H $-$ T$\Delta$S
where, $\Delta$G = change in Gibb's free energy
$\Delta$H = change in enthalpy
T = temperature
$\Delta$S = change in entropy
$\Delta$G = 57.8 kJ/mol $-$ [298 K $\times$ ($-$ 176 Jk$-$1 mol$-$1)]
= 57.8 kJ/mol $-$ $\left( {298 \times {{ - 176} \over {1000}}kJ} \right)$ [$\therefore$ 1 kJ = 1000 J]
= $-$ 5.352 kJ/mol
| $\Delta$G | = 5.352
Hence, answer is 5.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
Two flasks I and II shown below are connected by a valve of negligible volume.
When the valve is opened, the final pressure of the system in bar is x $\times$ 10$-$2. The value of x is __________. (Integer answer)
[Assume - Ideal gas; 1 bar = 105 Pa; Molar mass of N2 = 28.0 g mol$-$1; R = 8.31 J mol$-$1 K$-$1]
When the valve is opened, the final pressure of the system in bar is x $\times$ 10$-$2. The value of x is __________. (Integer answer)
[Assume - Ideal gas; 1 bar = 105 Pa; Molar mass of N2 = 28.0 g mol$-$1; R = 8.31 J mol$-$1 K$-$1]
Correct Answer: 84
Explanation:
Applying; (nI + nII)initial = (nI + nII)final
$\Rightarrow$ Assuming the system attains a final temperature of T (such that 300 < T < 60)
$\Rightarrow$ $\left( {\matrix{ {Heat\,lost\,by} \cr {{N_2}\,of\,container} \cr I \cr } } \right) = \left( {\matrix{ {Heat\,gained\,by} \cr {{N_2}\,of\,container} \cr {II} \cr } } \right)$
$\Rightarrow$ n1Cm(300 $-$ T) = nIICm(T $-$ 60)
$ \Rightarrow \left( {{{2.8} \over {28}}} \right)(300 - T) = {{0.2} \over {28}}(T - 60)$
$\Rightarrow$ 14(300 $-$ T) = T $-$ 60
$\Rightarrow$ ${{(14 \times 300 + 60)} \over {15}} = T$
$\Rightarrow$ T = 284 K (final temperature)
$\Rightarrow$ If the final pressure = P
$\Rightarrow$ (nI + nII)final = $\left( {{{3.0} \over {28}}} \right)$
$\Rightarrow$${P \over {RT}}({V_I} + {V_{II}}) = {{3.0\,gm} \over {28\,gm/mol}}$
$P = \left( {{3 \over {28}}mol} \right) \times 8.31{J \over {mol - K}} \times {{284K} \over {3 \times {{10}^{ - 3}}{m^3}}} \times {10^{ - 5}}{{bar} \over {Pa}}$
$\Rightarrow$ 0.84287 bar
$\Rightarrow$ 84.28 $\times$ 10$-$2 bar
$\Rightarrow$ 84
$\Rightarrow$ Assuming the system attains a final temperature of T (such that 300 < T < 60)
$\Rightarrow$ $\left( {\matrix{ {Heat\,lost\,by} \cr {{N_2}\,of\,container} \cr I \cr } } \right) = \left( {\matrix{ {Heat\,gained\,by} \cr {{N_2}\,of\,container} \cr {II} \cr } } \right)$
$\Rightarrow$ n1Cm(300 $-$ T) = nIICm(T $-$ 60)
$ \Rightarrow \left( {{{2.8} \over {28}}} \right)(300 - T) = {{0.2} \over {28}}(T - 60)$
$\Rightarrow$ 14(300 $-$ T) = T $-$ 60
$\Rightarrow$ ${{(14 \times 300 + 60)} \over {15}} = T$
$\Rightarrow$ T = 284 K (final temperature)
$\Rightarrow$ If the final pressure = P
$\Rightarrow$ (nI + nII)final = $\left( {{{3.0} \over {28}}} \right)$
$\Rightarrow$${P \over {RT}}({V_I} + {V_{II}}) = {{3.0\,gm} \over {28\,gm/mol}}$
$P = \left( {{3 \over {28}}mol} \right) \times 8.31{J \over {mol - K}} \times {{284K} \over {3 \times {{10}^{ - 3}}{m^3}}} \times {10^{ - 5}}{{bar} \over {Pa}}$
$\Rightarrow$ 0.84287 bar
$\Rightarrow$ 84.28 $\times$ 10$-$2 bar
$\Rightarrow$ 84
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
Data given for the following reaction is as follows :
FeO(s) + C(graphite) $\to$ Fe(s) + CO(g)
The minimum temperature in K at which the reaction becomes spontaneous is ___________. (Integer answer)
FeO(s) + C(graphite) $\to$ Fe(s) + CO(g)
| Substance | $\Delta H^\circ $ (kJ mol$^{ - 1}$) |
$\Delta S^\circ $ (J mol$^{ - 1}$ K$^{ - 1}$) |
|---|---|---|
| $Fe{O_{(s)}}$ | $ - 266.3$ | 57.49 |
| ${C_{(graphite)}}$ | 0 | 5.74 |
| $F{e_{(s)}}$ | 0 | 27.28 |
| $C{O_{(g)}}$ | $ - 110.5$ | 197.6 |
The minimum temperature in K at which the reaction becomes spontaneous is ___________. (Integer answer)
Correct Answer: 964
Explanation:
${T_{\min }} = \left( {{{{\Delta ^0}H} \over {{\Delta ^0}S}}} \right)$
${\Delta ^0}{H_{rxn}} = \left[ {\Delta _f^0H(Fe) + \Delta _f^0H(CO)} \right] - $
$ = \left[ {\Delta _f^0H(FeO) + \Delta _f^0H({C_{(graphite)}})} \right]$
$ = [0 - 110.5] - [ - 266.3 + 0]$
= 155.8 kJ/mol
${\Delta ^0}{S_{rxn}} = \left[ {{\Delta ^0}S(Fe) + {\Delta ^0}S(CO)} \right] - $
$\left[ {{\Delta ^0}S(FeO) + {\Delta ^0}S({C_{(graphite)}})} \right]$
$ = [27.28 + 197.6] - [57.49 + 5.74]$
= 161.65 J/mol-K
${T_{\min }} = {{155.8 \times {{10}^3}J/mol} \over {161.65J/mol - K}} = 963.8$ K
$ \simeq 964$ k (nearest integer)
${\Delta ^0}{H_{rxn}} = \left[ {\Delta _f^0H(Fe) + \Delta _f^0H(CO)} \right] - $
$ = \left[ {\Delta _f^0H(FeO) + \Delta _f^0H({C_{(graphite)}})} \right]$
$ = [0 - 110.5] - [ - 266.3 + 0]$
= 155.8 kJ/mol
${\Delta ^0}{S_{rxn}} = \left[ {{\Delta ^0}S(Fe) + {\Delta ^0}S(CO)} \right] - $
$\left[ {{\Delta ^0}S(FeO) + {\Delta ^0}S({C_{(graphite)}})} \right]$
$ = [27.28 + 197.6] - [57.49 + 5.74]$
= 161.65 J/mol-K
${T_{\min }} = {{155.8 \times {{10}^3}J/mol} \over {161.65J/mol - K}} = 963.8$ K
$ \simeq 964$ k (nearest integer)
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is $-$57.1 kJ. The increase in temperature in $^\circ$C of the system on mixing is x $\times$ 10$-$2. The value of x is ___________. (Nearest integer)
[Given : Specific heat of water = 4.18 J g$-$1 K$-$1, Density of water = 1.00 g cm$-$3]
[Assume no volume change on mixing)
[Given : Specific heat of water = 4.18 J g$-$1 K$-$1, Density of water = 1.00 g cm$-$3]
[Assume no volume change on mixing)
Correct Answer: 82
Explanation:
$\Rightarrow$ Millimoles of HCl = 200 $\times$ 0.2 = 40
$\Rightarrow$ Millimoles of NaOH = 300 $\times$ 0.1 = 30
$\Rightarrow$ Heat released = $\left( {{{30} \over {1000}} \times 57.1 \times 1000} \right)$ = 1713 J
$\Rightarrow$ Mass of solution = 500 ml $\times$ 1 gm/ml = 500 gm
$\Rightarrow$ $\Delta T = {q \over {m \times c}} = {{1713J} \over {500g \times 4.18{J \over {g - K}}}}$ = 0.8196 K
= 81.96 $\times$ 10$-$2 K
$\Rightarrow$ Millimoles of NaOH = 300 $\times$ 0.1 = 30
$\Rightarrow$ Heat released = $\left( {{{30} \over {1000}} \times 57.1 \times 1000} \right)$ = 1713 J
$\Rightarrow$ Mass of solution = 500 ml $\times$ 1 gm/ml = 500 gm
$\Rightarrow$ $\Delta T = {q \over {m \times c}} = {{1713J} \over {500g \times 4.18{J \over {g - K}}}}$ = 0.8196 K
= 81.96 $\times$ 10$-$2 K
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
For water $\Delta$vap H = 41 kJ mol$-$1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___________ kJ mol$-$1
[Use : R = 8.3 J mol$-$1 K$-$1]
[Use : R = 8.3 J mol$-$1 K$-$1]
Correct Answer: 38
Explanation:
H2O(l) $\to$ H2O(g) : $\Delta$H = 41 ${{kJ} \over {mol}}$
$\Rightarrow$ From the relation : $\Delta$H = $\Delta$U + $\Delta$ngRT
$\Rightarrow$ 41${{kJ} \over {mol}}$ = $\Delta$U + (1) $\times$ ${{8.3} \over {1000}}$ $\times$ 373
$\Delta$ DU = 41 $-$ 3.0959 = 38 kJ/mol
$\Rightarrow$ From the relation : $\Delta$H = $\Delta$U + $\Delta$ngRT
$\Rightarrow$ 41${{kJ} \over {mol}}$ = $\Delta$U + (1) $\times$ ${{8.3} \over {1000}}$ $\times$ 373
$\Delta$ DU = 41 $-$ 3.0959 = 38 kJ/mol
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Morning Shift
The Born-Haber cycle for KCl is evaluated with the following data :
${\Delta _f}{H^\Theta }$ for KCl = $-$436.7 kJ mol$-$1 ;
${\Delta _{sub}}{H^\Theta }$ for K = 89.2 kJ mol$-$1 ;
${\Delta _{ionization}}{H^\Theta }$ for K = 419.0 kJ mol$-$1 ;
${\Delta _{electron\,gain}}{H^\Theta }$ for Cl(g) = $-$348.6 kJ mol$-$1 ;
${\Delta _{bond}}{H^\Theta }$ for Cl2 = 243.0 kJ mol$-$1
The magnitude of lattice enthalpy of KCl in kJ mol$-$1 is _____________ (Nearest integer)
${\Delta _f}{H^\Theta }$ for KCl = $-$436.7 kJ mol$-$1 ;
${\Delta _{sub}}{H^\Theta }$ for K = 89.2 kJ mol$-$1 ;
${\Delta _{ionization}}{H^\Theta }$ for K = 419.0 kJ mol$-$1 ;
${\Delta _{electron\,gain}}{H^\Theta }$ for Cl(g) = $-$348.6 kJ mol$-$1 ;
${\Delta _{bond}}{H^\Theta }$ for Cl2 = 243.0 kJ mol$-$1
The magnitude of lattice enthalpy of KCl in kJ mol$-$1 is _____________ (Nearest integer)
Correct Answer: 718
Explanation:
${\Delta _f}H_{KCl}^\Theta = {\Delta _{sub}}H_{(K)}^\Theta + {\Delta _{ioniztion}}H_{(K)}^\Theta + {1 \over 2}{\Delta _{bond}}H_{(C{l_2})}^\Theta + {\Delta _{electron\,gain}}H_{(Cl)}^\Theta + {\Delta _{lattice}}H_{(KCl)}^\Theta $
$ \Rightarrow - 436.7 = 89.2 + 419.0 + {1 \over 2}(243.0) + \{ - 348.6\} + {\Delta _{lattice}}H_{(KCl)}^\Theta $
$ \Rightarrow {\Delta _{lattice}}H_{(KCl)}^\Theta = - 717.8$ kJ mol$-$1
The magnitude of lattice enthalpy of KCl in kJ mol$-$1 is 718 (Nearest integer).
$ \Rightarrow - 436.7 = 89.2 + 419.0 + {1 \over 2}(243.0) + \{ - 348.6\} + {\Delta _{lattice}}H_{(KCl)}^\Theta $
$ \Rightarrow {\Delta _{lattice}}H_{(KCl)}^\Theta = - 717.8$ kJ mol$-$1
The magnitude of lattice enthalpy of KCl in kJ mol$-$1 is 718 (Nearest integer).
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
When 400 mL of 0.2 M H2SO4 solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is __________ $\times$ 10$-$2 K. (Round off to the nearest integer).
[Use : H+ (aq) + OH$-$ (aq) $\to$ H2O : $\Delta$$\gamma$H = $-$57.1 kJ mol$-$1]
Specific heat of H2O = 4.18 J K$-$1 g$-$1
density of H2O = 1.0 g cm$-$3
Assume no change in volume of solution on mixing.
[Use : H+ (aq) + OH$-$ (aq) $\to$ H2O : $\Delta$$\gamma$H = $-$57.1 kJ mol$-$1]
Specific heat of H2O = 4.18 J K$-$1 g$-$1
density of H2O = 1.0 g cm$-$3
Assume no change in volume of solution on mixing.
Correct Answer: 82
Explanation:
${n_{{H^ + }}} = {{400 \times 0.2} \over {1000}} \times 2 = 0.16$
${n_{O{H^ - }}} = {{600 \times 0.1} \over {1000}} = 0.06$ (L.R.)
Now, heat liberated from reaction = heat gained by solutions
or, 0.06 $\times$ 57.1 $\times$ 103
= (1000 $\times$ 1.0) $\times$ 4.18 $\times$ $\Delta$T
$\therefore$ $\Delta$T = 0.8196K
= 81.96 $\times$ 10$-$2 K $ \approx $ 82 $\times$ 10$-$2 K
${n_{O{H^ - }}} = {{600 \times 0.1} \over {1000}} = 0.06$ (L.R.)
Now, heat liberated from reaction = heat gained by solutions
or, 0.06 $\times$ 57.1 $\times$ 103
= (1000 $\times$ 1.0) $\times$ 4.18 $\times$ $\Delta$T
$\therefore$ $\Delta$T = 0.8196K
= 81.96 $\times$ 10$-$2 K $ \approx $ 82 $\times$ 10$-$2 K
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
For water at 100$^\circ$ C and 1 bar,
$\Delta$vap H $-$ $\Delta$vap U = _____________ $\times$ 102 J mol$-$1. (Round off to the Nearest Integer)
[Use : R = 8.31 J mol$-$1 K$-$1]
[Assume volume of H2O(l) is much smaller than volume of H2O(g). Assume H2O(g) treated as an ideal gas]
$\Delta$vap H $-$ $\Delta$vap U = _____________ $\times$ 102 J mol$-$1. (Round off to the Nearest Integer)
[Use : R = 8.31 J mol$-$1 K$-$1]
[Assume volume of H2O(l) is much smaller than volume of H2O(g). Assume H2O(g) treated as an ideal gas]
Correct Answer: 31
Explanation:
H2O(l) $\rightleftharpoons$ H2O(v)
$\Delta$H = $\Delta$U + $\Delta$ngRT
For 1 mole waters;
$\Delta$ng = 1
$\therefore$ $\Delta$ngRT = 1 mol $\times$ 8.31 J/mol-k $\times$ 373 K
= 3099.63 J $ \cong $ 31 $\times$ 102 J
$\Delta$H = $\Delta$U + $\Delta$ngRT
For 1 mole waters;
$\Delta$ng = 1
$\therefore$ $\Delta$ngRT = 1 mol $\times$ 8.31 J/mol-k $\times$ 373 K
= 3099.63 J $ \cong $ 31 $\times$ 102 J
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
A system does 200 J of work and at the same time absorbs 150 J of heat. The magnitude of the change in internal energy is ____________ J. (Nearest integer)
Correct Answer: 50
Explanation:
w = $-$200 J, q = +150 : $\Delta$U = q + w
$\Delta$U = 150 $-$ 200 = $-$50 J
Magnitude = 50 J = |$\Delta$U |
$\Delta$U = 150 $-$ 200 = $-$50 J
Magnitude = 50 J = |$\Delta$U |
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
At 298 K, the enthalpy of fusion of a solid (X) is 2.8 kJ mol$-$1 and the enthalpy of vaporisation of the liquid (X) is 98.2 kJ mol$-$1. The enthalpy of sublimation of the substance (X) in kJ mol$-$1 is _____________. (in nearest integer)
Correct Answer: 101
Explanation:
The enthalpy of sublimation is the total amount of energy required to convert a solid directly into a gas. This can be calculated by summing the enthalpy of fusion (solid to liquid) and the enthalpy of vaporization (liquid to gas). Mathematically, this relationship is represented as:
$ \Delta H_{\text{sublimation}} = \Delta H_{\text{fusion}} + \Delta H_{\text{vaporization}} $
Given:
Enthalpy of fusion, $\Delta H_{\text{fusion}} = 2.8 \, \text{kJ mol}^{-1}$
Enthalpy of vaporization, $\Delta H_{\text{vaporization}} = 98.2 \, \text{kJ mol}^{-1}$
Substitute these values into the equation:
$ \Delta H_{\text{sublimation}} = 2.8 \, \text{kJ mol}^{-1} + 98.2 \, \text{kJ mol}^{-1} = 101.0 \, \text{kJ mol}^{-1} $
Therefore, the enthalpy of sublimation of the substance (X) is approximately 101 kJ mol$-1$.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
If the standard molar enthalpy change for combustion of graphite powder is $-$2.48 $\times$ 102 kJ mol$-$1, the amount of heat generated on combustion of 1 g of graphite powder is ___________ kJ. (Nearest integer)
Correct Answer: 21
Explanation:
1 mol graphite = 12 gm C
For 1 g of graphite = ${{248} \over {12}}$ = 20.67 kJ/gm heat evolved.
For 1 g of graphite = ${{248} \over {12}}$ = 20.67 kJ/gm heat evolved.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
For a given chemical reaction A $\to$ B at 300 K the free energy change is $-$49.4 kJ mol$-$1 and the enthalpy of reaction is 51.4 kJ mol$-$1. The entropy change of the reaction is _____________ JK$-$1 mol$-$1.
Correct Answer: 336
Explanation:
$\Delta$G = $-$49.4 kJ/mol
$\Delta$H = 51.4 kJ/mol
$\Delta$G = $\Delta$H $-$ T$\Delta$S
$-$49400 = 51400 $-$ 300$\Delta$S
$\Delta S = {{ + 100800} \over {300}} = 336$ JK$-$1 mol$-$1
$\Delta$H = 51.4 kJ/mol
$\Delta$G = $\Delta$H $-$ T$\Delta$S
$-$49400 = 51400 $-$ 300$\Delta$S
$\Delta S = {{ + 100800} \over {300}} = 336$ JK$-$1 mol$-$1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
For the reaction C2H6 $ \to $ C2H4 + H2
the reaction enthalpy $\Delta$rH = __________ kJ mol$-$1. (Round off to the Nearest Integer).
[ Given : Bond enthalpies in kJ mol$-$1 : C-C : 347, C = C : 611; C-H : 414, H-H : 436 ]
the reaction enthalpy $\Delta$rH = __________ kJ mol$-$1. (Round off to the Nearest Integer).
[ Given : Bond enthalpies in kJ mol$-$1 : C-C : 347, C = C : 611; C-H : 414, H-H : 436 ]
Correct Answer: 128
Explanation:
$\Delta H = {E_{C - C}} + 6{E_{C - H}} - {E_{C = C}} - 4{E_{C - H}} - {E_{H - H}}$
$ = 347 + 6(414) - 611 - 4 \times 414 - 436$
$ = 347 + 828 - 1047$
$ = 128$ KJ/ mol
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
The standard enthalpies of formation of Al2O3 and CaO are $-$1675 kJ mol-1 and $-$635 kJ mol$-$1 respectively.
For the reaction
3CaO + 2Al $ \to $ 3Ca + Al2O3 the standard reaction enthalpy $\Delta$rH0 = _________ kJ.
(Round off to the Nearest Integer)
For the reaction
3CaO + 2Al $ \to $ 3Ca + Al2O3 the standard reaction enthalpy $\Delta$rH0 = _________ kJ.
(Round off to the Nearest Integer)
Correct Answer: 230
Explanation:
$3CaO + 2Al\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{}} 3Ca + A{l_2}{O_3}$
$\Delta H_{reaction}^o = 3\Delta {H^o}_f(Ca,s) + \Delta {H^o}_f(A{l_2}{O_3},s) - 3\Delta {H^o}_f(CaO,s) - 2\Delta {H^o}_f(Al,S)$
$ = 0 + ( - 1675) - 3( - 635) - 0$
$ = - 1675 + 1905$
$ = 230$ KJ
$\Delta H_{reaction}^o = 3\Delta {H^o}_f(Ca,s) + \Delta {H^o}_f(A{l_2}{O_3},s) - 3\Delta {H^o}_f(CaO,s) - 2\Delta {H^o}_f(Al,S)$
$ = 0 + ( - 1675) - 3( - 635) - 0$
$ = - 1675 + 1905$
$ = 230$ KJ
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
At 25$^\circ$C, 50 g of iron reacts with HCl to form FeCl2. The evolved hydrogen gas expands against a constant pressure of 1 bar. The work done by the gas during this expansion is _________ J. (Round off to the Nearest Integer).
[Given : R = 8.314 J mol$-$1 K$-$1. Assume, hydrogen is an ideal gas] [Atomic mass of Fe is 55.85 u]
[Given : R = 8.314 J mol$-$1 K$-$1. Assume, hydrogen is an ideal gas] [Atomic mass of Fe is 55.85 u]
Correct Answer: 2218
Explanation:
$Fe + 2HCl \to FeC{l_2} + {H_2}$
${{50} \over {55.85}}moles$
Moles of Fe = ${{50} \over {55.85}}moles$ = Moles of H2
No. of H2 produced $ = {{50} \over {55.85}}moles$
Work done $ = - {P_{ext}}\,.\,\Delta V$
$ = - \Delta {n_g}RT$
$ = - {{50} \over {55.85}} \times 8.314 \times 298$
= -2218.05 J
Nearest integer = 2218
${{50} \over {55.85}}moles$
Moles of Fe = ${{50} \over {55.85}}moles$ = Moles of H2
No. of H2 produced $ = {{50} \over {55.85}}moles$
Work done $ = - {P_{ext}}\,.\,\Delta V$
$ = - \Delta {n_g}RT$
$ = - {{50} \over {55.85}} \times 8.314 \times 298$
= -2218.05 J
Nearest integer = 2218
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
The average S-F bond energy in kJ mol$-$1 of SF6 is _____________. (Rounded off to the nearest integer)
[Given : The values of standard enthalpy of formation of SF6(g), S(g) and F(g) are - 1100, 275 and 80 kJ mol$-$1 respectively.]
[Given : The values of standard enthalpy of formation of SF6(g), S(g) and F(g) are - 1100, 275 and 80 kJ mol$-$1 respectively.]
Correct Answer: 309
Explanation:
So, $\Delta_f H^{\circ}[S, g]+6 \times \Delta_f H^{\circ}[F, g]=\Delta_f H^{\circ}\left[SF_6, g\right]$ $+ 6 \times E_{S-F}$
$\left[\therefore E_{S-F}=\right.$ Average $S-\mathrm{F}$ bond energy in $\left.\mathrm{SF}_6\right]$
$275+6 \times 80=-1100+6 \times E_{S-F}$
$\Rightarrow E_{S-F}=\frac{275+6 \times 80+1100}{6}$ $=309.16 \mathrm{~kJ} \mathrm{~mol}^{-1}=309 \mathrm{~kJ} \mathrm{~mol} \mathrm{~m}^{-1}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
For a chemical reaction A + B ⇌ C + D
(${\Delta _r}{H^\Theta }$ = 80 kJ mol$-$1) the entropy change ${\Delta _r}{S^\Theta }$ depends on the temperature T (in K) as ${\Delta _r}{S^\Theta }$ = 2T (J K$-$1mol$-$1).
Minimum temperature at which it will become spontaneous is ___________ K. (Integer)
(${\Delta _r}{H^\Theta }$ = 80 kJ mol$-$1) the entropy change ${\Delta _r}{S^\Theta }$ depends on the temperature T (in K) as ${\Delta _r}{S^\Theta }$ = 2T (J K$-$1mol$-$1).
Minimum temperature at which it will become spontaneous is ___________ K. (Integer)
Correct Answer: 200
Explanation:
We know, $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ $
For a reaction to be spontaneous
$\Delta G^\circ < 0$
$ \Rightarrow $ $\Delta H^\circ - T\Delta S^\circ < 0$
$ \Rightarrow $ $T > {{\Delta H^\circ } \over {\Delta S^\circ }}$
$ \Rightarrow $ $T > {{80000} \over {2T}}$
$ \Rightarrow $ 2T2 > 80000
$ \Rightarrow $ T2 > 40000
$ \Rightarrow $ T > 200
The minimum temperature to make it spontaneous is 200 K.
For a reaction to be spontaneous
$\Delta G^\circ < 0$
$ \Rightarrow $ $\Delta H^\circ - T\Delta S^\circ < 0$
$ \Rightarrow $ $T > {{\Delta H^\circ } \over {\Delta S^\circ }}$
$ \Rightarrow $ $T > {{80000} \over {2T}}$
$ \Rightarrow $ 2T2 > 80000
$ \Rightarrow $ T2 > 40000
$ \Rightarrow $ T > 200
The minimum temperature to make it spontaneous is 200 K.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
An exothermic reaction X $ \to $ Y has an activation energy 30 kJ mol$-$1. If energy change $\Delta$E during the reaction is $-$20 kJ, then the activation energy for the reverse reaction in kJ is ___________. (Integer answer)
Correct Answer: 50
Explanation:
X $ \to $ Y
$\Delta $E = (Ea)f – (Ea)b
$ \Rightarrow $ – 20 = 30 – (Ea)b
$ \Rightarrow $ (Ea)b = 50 kJ
$\Delta $E = (Ea)f – (Ea)b
$ \Rightarrow $ – 20 = 30 – (Ea)b
$ \Rightarrow $ (Ea)b = 50 kJ
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is _________ kJ mol$-$1. (Rounded off to the nearest integer) [Use R = 8.314 J mol$-$1K$-$1]
Correct Answer: 15
Explanation:
The gas performs isothermal irreversible work (W).
where, $\Delta$U = 0 (change in internal energy)
From, 1st law of thermodynamics,
$\Rightarrow$ $\Delta$U = $\Delta$Q + W
$\Rightarrow$ 0 = $\Delta$Q + W
$\Rightarrow$ $\Delta$Q = $-$W
Now, $W = - {p_{ext}}({V_2} - {V_1})$
$ = - {p_{ext}}\left( {{{nRT} \over {{p_2}}} - {{nRT} \over {{p_1}}}} \right) = - {p_{ext}} \times nRT\left( {{1 \over {{p_2}}} - {1 \over {{p_1}}}} \right)$
Given, pext = 4.3 MPa, p1 = 2.1 MPa, p2 = 1.3 MPa, n = 5 mol, T = 293 K and R = 8.314 J mol$-$1 K$-$1
$ = - 4.3 \times 5 \times 8.314 \times 293\left( {{1 \over {1.3}} - {1 \over {2.1}}} \right)$
= $-$ 15347.70 J mol$-$1
= $-$ 15.347 kJ mol$-$1 $ \simeq $ $-$ 15 kJ mol$-$1
$\Rightarrow$ $\Delta$Q = 15 kJ mol$-$1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
The reaction of cyanamide, NH2CN(s) with oxygen was run in a bomb calorimeter and $\Delta$U was found to be $-$742.24 kJ mol$-$1. The magnitude of $\Delta$H298 for the reaction
$N{H_2}C{H_{(S)}} + {3 \over 2}{O_{2(g)}} \to {N_{2(g)}} + {O_{2(g)}} + {H_2}{O_{(I)}}$
is _________ kJ. (Rounded off to the nearest integer) [Assume ideal gases and R = 8.314 J mol$-$1 K$-$1]
$N{H_2}C{H_{(S)}} + {3 \over 2}{O_{2(g)}} \to {N_{2(g)}} + {O_{2(g)}} + {H_2}{O_{(I)}}$
is _________ kJ. (Rounded off to the nearest integer) [Assume ideal gases and R = 8.314 J mol$-$1 K$-$1]
Correct Answer: 741
Explanation:
$N{H_2}CN(s) + {3 \over 2}{O_2}(g)\buildrel {} \over
\longrightarrow {N_2}(g) + C{O_2}(g) + {H_2}O(l)$
$\Delta ng = (1 + 1) - {3 \over 2} = {1 \over 2}$
$\Delta H = \Delta U + \Delta ng\,RT$
$ = - 742.24 + {1 \over 2} \times {{8.314 \times 298} \over {1000}}$
$ = - 742.24 + 1.24$
$ = -741$ kJ/mol
$\Delta ng = (1 + 1) - {3 \over 2} = {1 \over 2}$
$\Delta H = \Delta U + \Delta ng\,RT$
$ = - 742.24 + {1 \over 2} \times {{8.314 \times 298} \over {1000}}$
$ = - 742.24 + 1.24$
$ = -741$ kJ/mol
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
The ionization enthalpy of Na+ formation from Na(g) is 495.8 kJ mol$-$1, while the electron gain enthalpy of Br is $-$325.0 kJ mol$-$1. Given the lattice enthalpy of NaBr is $-$728.4 kJ mol$-$1. The energy for the formation of NaBr ionic solid is ($-$) ____________ $\times$ 10$-$1 kJ mol$-$1.
Correct Answer: 5576
Explanation:
$Na(s)\buildrel {} \over
\longrightarrow N{a^ + }(g)$
$\Delta H = 495.8$
${1 \over 2}B{r_2}(l) + {e^ - }\buildrel {} \over \longrightarrow B{r^ - }(g)$
$\Delta H = 325$
$N{a^ + }(g) + B{r^ - }(g)\buildrel {} \over \longrightarrow NaBr(s)$
$\Delta H = - 728.4$
$Na(s) + {1 \over 2}B{r_2}(l)\buildrel {} \over \longrightarrow NaBr(s).$
$\Delta H = ?$
$\Delta H = 495.8 - 325 - 728.4 - 557.6$ kJ
$ = - 5576 \times {10^{ - 1}}$ kJ
$\Delta H = 495.8$
${1 \over 2}B{r_2}(l) + {e^ - }\buildrel {} \over \longrightarrow B{r^ - }(g)$
$\Delta H = 325$
$N{a^ + }(g) + B{r^ - }(g)\buildrel {} \over \longrightarrow NaBr(s)$
$\Delta H = - 728.4$
$Na(s) + {1 \over 2}B{r_2}(l)\buildrel {} \over \longrightarrow NaBr(s).$
$\Delta H = ?$
$\Delta H = 495.8 - 325 - 728.4 - 557.6$ kJ
$ = - 5576 \times {10^{ - 1}}$ kJ
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
Assuming ideal behaviour, the magnitude of log K for the following reaction at 25$^\circ$C is x $\times$ 10$-$1. The value of x is ______________. (Integer answer)
$3HC \equiv C{H_{(g)}} \rightleftharpoons {C_6}{H_{6(l)}}$
[Given : ${\Delta _f}{G^o}(HC \equiv CH) = - 2.04 \times {10^5}$ J mol$-$1 ; ${\Delta _f}{G^o}({C_6}{H_6}) = - 1.24 \times {10^5}$ J mol$-$1 ; R = 8.314 J K-1 mol$-$1]
$3HC \equiv C{H_{(g)}} \rightleftharpoons {C_6}{H_{6(l)}}$
[Given : ${\Delta _f}{G^o}(HC \equiv CH) = - 2.04 \times {10^5}$ J mol$-$1 ; ${\Delta _f}{G^o}({C_6}{H_6}) = - 1.24 \times {10^5}$ J mol$-$1 ; R = 8.314 J K-1 mol$-$1]
Correct Answer: 855
Explanation:
Reaction,
$\mathop {3HC \equiv CH(g)}\limits_{Acetylene} \to \mathop {{C_6}{H_6}(l)}\limits_{Benzene} $
Given, $\Delta G_f^o (CH \equiv CH) = - 2.04 \times {10^5}$ J mol$-$1
$\Delta G_f^o({C_6}{H_6}) = - 1.24 \times {10^5}$ J mol$-$1
Gibb's free energy, $\Delta G_f^o = - nRT\ln K$
$\Delta G_f^o = \sum {{{(\Delta G_F^o)}_P} - \sum {{{(\Delta G_F^o)}_R}} } $
$ - nRT\ln K = - n'RT\ln {K_p} - ( - n''RT\ln {K_f})$
$ \Rightarrow - RT\ln K = 1 \times ( - 1.24 \times {10^5}) - ( - 3 \times 2.04 \times {10^5}) $
$ \Rightarrow $ $- 2.303 \times R \times T\log K = 4.88 \times {10^5}$
$ \Rightarrow $ $\log K = - {{4.88 \times {{10}^5}} \over {2.303 \times 8.314 \times 273}}$
$ \Rightarrow $ $n\ln K = n'\ln {K_p} - ( - n''ln{K_f})$
$ \Rightarrow $ K = 85.52
$\Rightarrow$ K = 855 $\times$ 10$-$1
x = 855
$\mathop {3HC \equiv CH(g)}\limits_{Acetylene} \to \mathop {{C_6}{H_6}(l)}\limits_{Benzene} $
Given, $\Delta G_f^o (CH \equiv CH) = - 2.04 \times {10^5}$ J mol$-$1
$\Delta G_f^o({C_6}{H_6}) = - 1.24 \times {10^5}$ J mol$-$1
Gibb's free energy, $\Delta G_f^o = - nRT\ln K$
$\Delta G_f^o = \sum {{{(\Delta G_F^o)}_P} - \sum {{{(\Delta G_F^o)}_R}} } $
$ - nRT\ln K = - n'RT\ln {K_p} - ( - n''RT\ln {K_f})$
$ \Rightarrow - RT\ln K = 1 \times ( - 1.24 \times {10^5}) - ( - 3 \times 2.04 \times {10^5}) $
$ \Rightarrow $ $- 2.303 \times R \times T\log K = 4.88 \times {10^5}$
$ \Rightarrow $ $\log K = - {{4.88 \times {{10}^5}} \over {2.303 \times 8.314 \times 273}}$
$ \Rightarrow $ $n\ln K = n'\ln {K_p} - ( - n''ln{K_f})$
$ \Rightarrow $ K = 85.52
$\Rightarrow$ K = 855 $\times$ 10$-$1
x = 855
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
For a reaction,
4M(s) + nO2(g) $ \to $ 2M2On(s)
the free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which :
4M(s) + nO2(g) $ \to $ 2M2On(s)
the free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which :
A.
the free energy change shows a change
from negative to positive value
B.
the slope changes from positive to negative
C.
the slope changes from negative to positive
D.
the slope changes from positive to zero
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol–1, and 4 kJ mol–1, respectively.
The hydration enthalpy of NaCl is :
The hydration enthalpy of NaCl is :
A.
–780 kJ mol–1
B.
–784 kJ mol–1
C.
780 kJ mol–1
D.
784 kJ mol–1
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
Five moles of an ideal gas at 1 bar and 298 K
is expanded into vacuum to double the volume.
The work done is :
A.
Zero
B.
-RT $\ln {{{V_2}} \over {{V_1}}}$
C.
CV (T2 – T1)
D.
– RT (V2 – V1)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
The process that is NOT endothermic in nature
is :
A.
Ar(g) + e- $ \to $ Ar-(g)
B.
H(g) + e- $ \to $ H-(g)
C.
Na(g) $ \to $ Na+(g) + e-
D.
O-(g) + e- $ \to $ O2-(g)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
For one mole of an ideal gas, which of these
statements must be true?
(a) U and H each depends only on temperature
(b) Compressibility factor z is not equal to 1
(c) CP, m – CV, m = R
(d) dU = CVdT for any process
(a) U and H each depends only on temperature
(b) Compressibility factor z is not equal to 1
(c) CP, m – CV, m = R
(d) dU = CVdT for any process
A.
(a), (c) and (d)
B.
(a) and (c)
C.
(c) and (d)
D.
(b), (c) and (d)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
The true statement amongst the following is :
A.
S is a function of temperature but $\Delta $S is not
a function of temperature.
B.
Both S and $\Delta $S are not functions of
temperature.
C.
Both $\Delta $S and S are functions of temperature.
D.
S is not a function of temperature but $\Delta $S is
a function of temperature.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
If enthalpy of atomisation for Br2(1) is x kJ/mol
and bond enthalpy for Br2 is y kJ/mol, the
relation between them :
A.
does not exist
B.
is x < y
C.
is x > y
D.
is x = y
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 5th September Evening Slot
For a dimerization reaction,
2A(g) $ \to $ A2(g)
at 298 K, $\Delta $Uo = –20 kJ mol–1, $\Delta $So = –30 JK–1 mol–1,
then the $\Delta $Go will be _____ J.
2A(g) $ \to $ A2(g)
at 298 K, $\Delta $Uo = –20 kJ mol–1, $\Delta $So = –30 JK–1 mol–1,
then the $\Delta $Go will be _____ J.
Correct Answer: -13540to-13537
Explanation:
$\Delta $Go = $\Delta $Ho - T$\Delta $So
= ($\Delta $Uo + $\Delta $ngRT) - T$\Delta $So
= $\left[ {\left\{ { - 20 + \left( 1 \right){{8.314} \over {1000}} \times 298} \right\} - {{298} \over {1000}} \times \left( { - 30} \right)} \right]$ kJ
= – 13.537572 kJ
= – 13537.57 Joule
= ($\Delta $Uo + $\Delta $ngRT) - T$\Delta $So
= $\left[ {\left\{ { - 20 + \left( 1 \right){{8.314} \over {1000}} \times 298} \right\} - {{298} \over {1000}} \times \left( { - 30} \right)} \right]$ kJ
= – 13.537572 kJ
= – 13537.57 Joule
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Evening Slot
The heat of combustion of ethanol into carbon
dioxide and water is – 327 kcal at constant
pressure. The heat evolved (in cal) at constant
volume and 27oC (if all gases behave ideally) is
(R = 2 cal mol–1 K–1) ________.
Correct Answer: 326400
Explanation:
C2H5OH (l) + 3O2(g) $ \to $ 2CO2(g) + 3H2O(l)
$\Delta $H = –327 kcal; $\Delta $ng = – 1
$\Delta $H = $\Delta $U + $\Delta $ngRT
$ \Rightarrow $ $\Delta $U = – 327 + 2 × 10–3 × 300
= – 326.4 kcal
= – 326400 cal
$ \therefore $ Heat evolved = 326400 cal.
$\Delta $H = –327 kcal; $\Delta $ng = – 1
$\Delta $H = $\Delta $U + $\Delta $ngRT
$ \Rightarrow $ $\Delta $U = – 327 + 2 × 10–3 × 300
= – 326.4 kcal
= – 326400 cal
$ \therefore $ Heat evolved = 326400 cal.
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Morning Slot
The internal energy change (in J) When 90 g of
water undergoes complete evaporation at
100oC is ____________.
(Given : $\Delta $Hvap for water at 373 K = 41 kJ/mol,
R = 8.314 JK–1 mol–1)
(Given : $\Delta $Hvap for water at 373 K = 41 kJ/mol,
R = 8.314 JK–1 mol–1)
Correct Answer: 189494TO189495
Explanation:
H2O(l) ⇌ H2O(g)
90 gm of H2O = ${{90} \over {18}}$ moles of H2O = 5 moles of H2O
$\Delta $Hvap = $\Delta $U + $\Delta $ngRT
$ \Rightarrow $ $\Delta $U = $\Delta $Hvap - $\Delta $ngRT
= 41000 - 1$ \times $8.314$ \times $373
= 37898.878
For 5 moles, $\Delta $U = 37898.878 $ \times $ 5 = 189494.39 Joule
90 gm of H2O = ${{90} \over {18}}$ moles of H2O = 5 moles of H2O
$\Delta $Hvap = $\Delta $U + $\Delta $ngRT
$ \Rightarrow $ $\Delta $U = $\Delta $Hvap - $\Delta $ngRT
= 41000 - 1$ \times $8.314$ \times $373
= 37898.878
For 5 moles, $\Delta $U = 37898.878 $ \times $ 5 = 189494.39 Joule
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Evening Slot
At constant volume, 4 mol of an ideal gas when
heated from 300 K to 500K changes its internal
energy by 5000 J. The molar heat capacity at
constant volume is _______.
Correct Answer: 6.25
Explanation:
$\Delta $U = nCv$\Delta $T
$ \Rightarrow $ 5000 = 4 × Cv(500 – 300)
$ \Rightarrow $ Cv = 6.25 JK–1mol–1
$ \Rightarrow $ 5000 = 4 × Cv(500 – 300)
$ \Rightarrow $ Cv = 6.25 JK–1mol–1
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Morning Slot
The magnitude of work done by a gas that undergoes a reversible expansion along the path ABC shown in
the figure is _______.
Correct Answer: 48
Explanation:
Work done = Area covered by the diagram
= $1\over2$ × (sum of parallel sides) × height
= $1\over2$ × (10+6) × 6
= $1\over2$ × 16 × 6
= 48 Joule
Note : Here pressure difference = 8 - 2 = 6 Pa and volume difference = 12 - 2 = 10
= $1\over2$ × (sum of parallel sides) × height
= $1\over2$ × (10+6) × 6
= $1\over2$ × 16 × 6
= 48 Joule
Note : Here pressure difference = 8 - 2 = 6 Pa and volume difference = 12 - 2 = 10
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Evening Slot
The standard heat of formation $\left( {{\Delta _f}H_{298}^0} \right)$ of ethane (in kj/mol), if the heat of combustion of
ethane, hydrogen and graphite are - 1560, -393.5 and -286 Kj/mol, respectively is :
Correct Answer: -192.5
Explanation:
2C(graphite)+ 3H(g) $ \to $ C2H6(g)
${\Delta _f}H$(C2H6) = 2$\Delta H$comb(Cgraphite) + 3$\Delta H$comb(H2)
- $\Delta H$comb(C2H6)
= – (286$ \times $2) - (393.5$ \times $3) -(-1560)
= -572 –1180.5+1560 = -192.5 kJ/mole
${\Delta _f}H$(C2H6) = 2$\Delta H$comb(Cgraphite) + 3$\Delta H$comb(H2)
- $\Delta H$comb(C2H6)
= – (286$ \times $2) - (393.5$ \times $3) -(-1560)
= -572 –1180.5+1560 = -192.5 kJ/mole
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Morning Slot
For the reaction :
A($l$) $ \to $ 2B(g)
$\Delta U = 2.1\,kcal,\,\Delta S = 20\,cal\,{K^{ - 1}}$ at 300 K
Hence $\Delta $G in kcal is :
A($l$) $ \to $ 2B(g)
$\Delta U = 2.1\,kcal,\,\Delta S = 20\,cal\,{K^{ - 1}}$ at 300 K
Hence $\Delta $G in kcal is :
Correct Answer: -2.7
Explanation:
A($l$) $ \to $ 2B(g)
We know, $\Delta $H = $\Delta $U + $\Delta $ngRT
and $\Delta $G = $\Delta $H - T$\Delta $S
$ \therefore $ $\Delta $G = $\Delta $U + $\Delta $ngRT - T$\Delta $S
= 2.1 + ${{2 \times 2 \times 300} \over {1000}}$ - ${{300 \times 20} \over {1000}}$
= 2.1 + 1.2 - 6 = – 2.70 Kcal/mol
We know, $\Delta $H = $\Delta $U + $\Delta $ngRT
and $\Delta $G = $\Delta $H - T$\Delta $S
$ \therefore $ $\Delta $G = $\Delta $U + $\Delta $ngRT - T$\Delta $S
= 2.1 + ${{2 \times 2 \times 300} \over {1000}}$ - ${{300 \times 20} \over {1000}}$
= 2.1 + 1.2 - 6 = – 2.70 Kcal/mol