Thermodynamics
354 Questions
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
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 2 Online
One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are same, the value of $\ln {{{V_3}} \over {{V_2}}}$ is _________.
(U : internal energy, S : entropy, p : pressure, V : volume, R : gas constant)
(Given : molar heat capacity at constant volume, CV,m of the gas is ${5 \over 2}$R)
(U : internal energy, S : entropy, p : pressure, V : volume, R : gas constant)
(Given : molar heat capacity at constant volume, CV,m of the gas is ${5 \over 2}$R)
Correct Answer: 10
Explanation:
For process 1, entropy is constant thus, q is constant
$\therefore$ ${W_I} = \Delta U = n{C_{V,m}}\Delta T$
$ = - (2250 - 450)R = - 1800R$ .... (i)
$\Delta U = n{C_{V,m}}\Delta T$
$ - 1800R = 1 \times {{5R} \over 2} \times \Delta T$
$ \Rightarrow \Delta T = - 720K$
${T_2} - {T_1} = - 720K$
${T_2} = - 720K + 900K = 180K$
For process II,
${W_{II}} = - nR{T_2}\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$
$ = - 1 \times R \times 180\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$ .... (ii)
As, both work done for process I and II are equal,
Therefore, WI = WII
$ - 1800R = - R \times 180 \times \ln \left( {{{{V_3}} \over {{V_2}}}} \right)$
$\ln \left( {{{{V_3}} \over {{V_2}}}} \right) = {{1800} \over {180}} \Rightarrow \ln \left( {{{{V_3}} \over {{V_2}}}} \right) = 10$
$\therefore$ ${W_I} = \Delta U = n{C_{V,m}}\Delta T$
$ = - (2250 - 450)R = - 1800R$ .... (i)
$\Delta U = n{C_{V,m}}\Delta T$
$ - 1800R = 1 \times {{5R} \over 2} \times \Delta T$
$ \Rightarrow \Delta T = - 720K$
${T_2} - {T_1} = - 720K$
${T_2} = - 720K + 900K = 180K$
For process II,
${W_{II}} = - nR{T_2}\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$
$ = - 1 \times R \times 180\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$ .... (ii)
As, both work done for process I and II are equal,
Therefore, WI = WII
$ - 1800R = - R \times 180 \times \ln \left( {{{{V_3}} \over {{V_2}}}} \right)$
$\ln \left( {{{{V_3}} \over {{V_2}}}} \right) = {{1800} \over {180}} \Rightarrow \ln \left( {{{{V_3}} \over {{V_2}}}} \right) = 10$
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 1 Online
The value of standard enthalpy, $\Delta$Ho (in kJ mol$-$1) for the given reaction is _______.
Correct Answer: 166.28
Explanation:
For the given reaction,
$ \mathrm{X}(\mathrm{s}) \rightleftharpoons \mathrm{X}(\mathrm{s})+\mathrm{Z}(\mathrm{g}) $
We have
$ \frac{d(\ln K)}{d(1 / T)}=\frac{-\Delta H^{\circ}}{R} $
And $ K=\frac{p_{z}}{p^{\circ}} $
Substituting the value of $K$ in Eq. (1), we get
$ \frac{d\left(\ln \frac{p_{z}}{p^{\circ}}\right)}{d\left(\frac{1}{T}\right)}=\frac{-\Delta H^{\circ}}{R} $
Or
$ d\left(\ln \frac{p_{z}}{p^{\circ}}\right)=\frac{-\Delta H^{\circ}}{R} d\left(\frac{1}{T}\right) $
Integrating both sides of Eq. (2), we get
$ \ln \frac{p_{z}}{p^{\circ}}=\frac{-\Delta H^{\circ}}{R T}+\ln A $
Where $\ln A=$ Integration constant
From the given plot of $\ln \frac{p_{z}}{p^{\circ}} \mathrm{vs} \frac{10^{4}}{T}$
The slope of the plot is
$ \text { Slope }=\frac{-\Delta H^{\circ}}{R}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-7-(-3)}{\left(\frac{12}{10^{4}}-\frac{10}{10^{4}}\right)} $
$\frac{-\Delta H^{\circ}}{R}=\frac{-4}{2} \times 10^{4}$
$\Delta H^{\circ}=-2 \times 10^{4} \times R$
$\Delta H^{\circ}=2 \times 8.314 \times 10^{4}$
$\Delta H^{\circ}=16.628 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}$
$\Delta H^{\circ}=166.28 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$ \mathrm{X}(\mathrm{s}) \rightleftharpoons \mathrm{X}(\mathrm{s})+\mathrm{Z}(\mathrm{g}) $
We have
$ \frac{d(\ln K)}{d(1 / T)}=\frac{-\Delta H^{\circ}}{R} $
And $ K=\frac{p_{z}}{p^{\circ}} $
Substituting the value of $K$ in Eq. (1), we get
$ \frac{d\left(\ln \frac{p_{z}}{p^{\circ}}\right)}{d\left(\frac{1}{T}\right)}=\frac{-\Delta H^{\circ}}{R} $
Or
$ d\left(\ln \frac{p_{z}}{p^{\circ}}\right)=\frac{-\Delta H^{\circ}}{R} d\left(\frac{1}{T}\right) $
Integrating both sides of Eq. (2), we get
$ \ln \frac{p_{z}}{p^{\circ}}=\frac{-\Delta H^{\circ}}{R T}+\ln A $
Where $\ln A=$ Integration constant
From the given plot of $\ln \frac{p_{z}}{p^{\circ}} \mathrm{vs} \frac{10^{4}}{T}$
The slope of the plot is
$ \text { Slope }=\frac{-\Delta H^{\circ}}{R}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-7-(-3)}{\left(\frac{12}{10^{4}}-\frac{10}{10^{4}}\right)} $
$\frac{-\Delta H^{\circ}}{R}=\frac{-4}{2} \times 10^{4}$
$\Delta H^{\circ}=-2 \times 10^{4} \times R$
$\Delta H^{\circ}=2 \times 8.314 \times 10^{4}$
$\Delta H^{\circ}=16.628 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}$
$\Delta H^{\circ}=166.28 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 1 Online
The value of $\Delta$S$\theta$ (in J K$-$1 mol$-$1) for the given reaction, at 1000 K is _________.
Correct Answer: 141.34
Explanation:
For the given reaction,
$ \mathrm{X}(\mathrm{s}) \rightleftharpoons \mathrm{X}(\mathrm{s})+\mathrm{Z}(\mathrm{g}) $
We have
$ \frac{d(\ln K)}{d(1 / T)}=\frac{-\Delta H^{\circ}}{R} $
And $ K=\frac{p_{z}}{p^{\circ}} $
Substituting the value of $K$ in Eq. (1), we get
$ \frac{d\left(\ln \frac{p_{z}}{p^{\circ}}\right)}{d\left(\frac{1}{T}\right)}=\frac{-\Delta H^{\circ}}{R} $
Or
$ d\left(\ln \frac{p_{z}}{p^{\circ}}\right)=\frac{-\Delta H^{\circ}}{R} d\left(\frac{1}{T}\right) $
Integrating both sides of Eq. (2), we get
$ \ln \frac{p_{z}}{p^{\circ}}=\frac{-\Delta H^{\circ}}{R T}+\ln A $
Where $\ln A=$ Integration constant
We can also calculate the intercept $(\ln A)$ from the plot of $\ln \frac{p_{z}}{p^{\circ}} \mathrm{vs} \frac{1}{T}$.
Now, from the equation of straight line, we have
or
$ \begin{aligned} &y=m x+C \\ &C=y-m x \end{aligned} $
From the plot, we have
$ m=-2 \times 10^{4} \mathrm{~K}, y=-3, x=\frac{10}{10^{4}} \mathrm{~K} $
Now, substituting the above values in Eq. (4), we get
$ \begin{aligned} &C=-3-\left(-2 \times 10^{4} \times \frac{10}{10^{4}}\right) \\ &C=-3+20 \\ &C=17 \end{aligned} $
$\Rightarrow C=17$
This intercept is equal to $\ln A$ in Eq. (3), so we have
$ \ln A=C=17 $
From the thermodynamic equations, we have
$ \Delta G^{\circ}=\Delta H^{\circ}-7 \Delta S^{\circ} $
For any equilibrium reaction, the Gibb's energy is,
$ \Delta G^{\circ}=-R T \ln K $
On comparing both the equations, we get
$ -R T \ln K=\Delta H^{\circ}-T \Delta S^{\circ} $
Or
$ \begin{aligned} \ln K &=\frac{-\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R} \\ \left(\frac{p_{z}}{p^{\circ}}\right) &=\frac{-\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R} \end{aligned} $
On comparing Eq. (3) and (6), we have
$ \frac{\Delta S^{\circ}}{R}=\ln A=17 $
Or
$ \begin{aligned} \Delta S^{\circ} &=17 R \\ &=17 \times 8.314 \\ &=141.34 \,\mathrm{JK}^{-1} \end{aligned} $
$ \mathrm{X}(\mathrm{s}) \rightleftharpoons \mathrm{X}(\mathrm{s})+\mathrm{Z}(\mathrm{g}) $
We have
$ \frac{d(\ln K)}{d(1 / T)}=\frac{-\Delta H^{\circ}}{R} $
And $ K=\frac{p_{z}}{p^{\circ}} $
Substituting the value of $K$ in Eq. (1), we get
$ \frac{d\left(\ln \frac{p_{z}}{p^{\circ}}\right)}{d\left(\frac{1}{T}\right)}=\frac{-\Delta H^{\circ}}{R} $
Or
$ d\left(\ln \frac{p_{z}}{p^{\circ}}\right)=\frac{-\Delta H^{\circ}}{R} d\left(\frac{1}{T}\right) $
Integrating both sides of Eq. (2), we get
$ \ln \frac{p_{z}}{p^{\circ}}=\frac{-\Delta H^{\circ}}{R T}+\ln A $
Where $\ln A=$ Integration constant
We can also calculate the intercept $(\ln A)$ from the plot of $\ln \frac{p_{z}}{p^{\circ}} \mathrm{vs} \frac{1}{T}$.
Now, from the equation of straight line, we have
or
$ \begin{aligned} &y=m x+C \\ &C=y-m x \end{aligned} $
From the plot, we have
$ m=-2 \times 10^{4} \mathrm{~K}, y=-3, x=\frac{10}{10^{4}} \mathrm{~K} $
Now, substituting the above values in Eq. (4), we get
$ \begin{aligned} &C=-3-\left(-2 \times 10^{4} \times \frac{10}{10^{4}}\right) \\ &C=-3+20 \\ &C=17 \end{aligned} $
$\Rightarrow C=17$
This intercept is equal to $\ln A$ in Eq. (3), so we have
$ \ln A=C=17 $
From the thermodynamic equations, we have
$ \Delta G^{\circ}=\Delta H^{\circ}-7 \Delta S^{\circ} $
For any equilibrium reaction, the Gibb's energy is,
$ \Delta G^{\circ}=-R T \ln K $
On comparing both the equations, we get
$ -R T \ln K=\Delta H^{\circ}-T \Delta S^{\circ} $
Or
$ \begin{aligned} \ln K &=\frac{-\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R} \\ \left(\frac{p_{z}}{p^{\circ}}\right) &=\frac{-\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R} \end{aligned} $
On comparing Eq. (3) and (6), we have
$ \frac{\Delta S^{\circ}}{R}=\ln A=17 $
Or
$ \begin{aligned} \Delta S^{\circ} &=17 R \\ &=17 \times 8.314 \\ &=141.34 \,\mathrm{JK}^{-1} \end{aligned} $
2021
JEE Advanced
MSQ
JEE Advanced 2021 Paper 1 Online
An ideal gas undergoes a reversible isothermal expansion from state I to state II followed by a reversible adiabatic expansion from state II to state III. The correct plot(s) representing the changes from state I to state III is (are)
(p : pressure, V : volume, T : temperature, H : enthalpy, S : entropy)
(p : pressure, V : volume, T : temperature, H : enthalpy, S : entropy)
A.
B.
C.
D.
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Evening Shift
When an ideal gas expands isothermally from $5 \mathrm{~m}^3$ to $10 \mathrm{~m}^3$ at $25^{\circ} \mathrm{C}$ against a constant pressure of $10^7 \mathrm{~Nm}^{-2}$, then the work done on the gas is
A.
$-100 \mathrm{~MJ}$
B.
$-50 \mathrm{~MJ}$
C.
$-0.5 \mathrm{~MJ}$
D.
$-10^5 \mathrm{~MJ}$
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Evening Shift
Find the approximate value of $(\Delta H-\Delta U)$ in $\mathrm{Jmol}^{-1}$, for the formation of CO from its elements at $298 \mathrm{~K} .\left(R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$
A.
$-1238$
B.
1238
C.
2477
D.
$-2477$
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Morning Shift
For the reaction, $\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ at $T=100^{\circ} \mathrm{C}$ and $p=1 \mathrm{~atm}$, choose the correct option.
A.
$\Delta S_{\text {system }} > 0$ and $\Delta S_{\text {surrounding }} > 0$
B.
$\Delta S_{\text {system }} > 0$ and $\Delta S_{\text {surrounding }} < 0$
C.
$\Delta S_{\text {system }} < 0$ and $\Delta S_{\text {Surrounding }} > 0$
D.
$\Delta S_{\text {system }} < 0$ and $\Delta S_{\text {surrounding }} < 0$
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Evening Shift
Two flasks $A$ and $B$ have equal volumes. $A$ is maintained at $300 \mathrm{~K}$ and $B$ at $600 \mathrm{~K}$. Equal masses of $\mathrm{H}_2$ and $\mathrm{CO}_2$ are taken in flasks $A$ and $B$ respectively. Find the ratio of total KE of gases in flask $A$ to that of $B$.
A.
1 : 2
B.
11 : 1
C.
33 : 2
D.
55 : 7
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Evening Shift
When the temperature of 2 moles of an ideal gas is increased by 20$^\circ$C at constant pressure. Find the work involved in the process.
A.
5R
B.
40R
C.
15R
D.
20R
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Morning Shift
If a chemical reaction is known to be non-spontaneous at 298 K but spontaneous at 350 K, then which among the following conditions is true for the reaction?
A.
$\Delta G=-v e, \Delta H=-v e, \Delta S=+ ve$
B.
$\Delta G=+v e, \Delta H=+v e, \Delta S=+v e$
C.
$\Delta G=-v e, \Delta H=+v e, \Delta S=+v e$
D.
$\Delta G=+ ve, \Delta H=+ ve, \Delta S=-v e$
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
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K) at which the reduction of cassiterite by coke would take place.
At $298K:{\Delta _f}H^\circ [Sn{O_2}(s)] = - 581.0$ mol-1,
$\eqalign{ & {\Delta _f}H^\circ [(C{O_2})(g)] = - 394.0\,kJ\,mol{ ^{-1}} \cr & S^\circ [Sn{O_2}(s)] = 56.0J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [Sn(s)] = 52.0\,J\,K{ ^{-1}}mo{l^{ - 1}} \cr & S^\circ [C(s)] = 6.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [C{O_2}(g)] = 210.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr} $
Assume that, the enthalpies and the entropies are temperature independent.
At $298K:{\Delta _f}H^\circ [Sn{O_2}(s)] = - 581.0$ mol-1,
$\eqalign{ & {\Delta _f}H^\circ [(C{O_2})(g)] = - 394.0\,kJ\,mol{ ^{-1}} \cr & S^\circ [Sn{O_2}(s)] = 56.0J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [Sn(s)] = 52.0\,J\,K{ ^{-1}}mo{l^{ - 1}} \cr & S^\circ [C(s)] = 6.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [C{O_2}(g)] = 210.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr} $
Assume that, the enthalpies and the entropies are temperature independent.
Correct Answer: 935
Explanation:
Tin is obtained from cassiterite by reduction with coke and the balanced chemical reaction is
$Sn{O_2}(s) + C(s)\buildrel {} \over \longrightarrow Sn(s) + C{O_2}(g)$
Standard enthalpy of reaction,
$\Delta H_{{R^\pi }}^o = (\Delta H_f^oC{O_2}(g)) - (\Delta H_f^oSn{O_2}(s))$
$\Delta H_{{R^\pi }}^o = - 394 - ( - 581) \Rightarrow 187$
Standard entropy of reaction,
$\Delta S_{{R^\pi }}^o = \Delta S_{\Pr oducts}^o - \Delta S_{{\mathop{\rm Re}\nolimits} ac\tan ts}^o$
$\Delta S_{{R^\pi }}^o = [S_{}^o(Sn(s)) + S_{}^o(C{O_2}(g))] - S_{}^o(Sn{O_2}(s)) + S_{}^o(C(s))]$
$\Delta S_{{R^\pi }}^o = [52 + 210] - [56 + 6]$
$ \Rightarrow $ 200 JK$-$1 mol$-$1
We know that, $\Delta H^\circ = T\Delta S^\circ $
$ \therefore $ $T = {{\Delta H^\circ } \over {\Delta S^\circ }}$
$ = {{187 \times 1000} \over {200}} \Rightarrow 935K$
For the reaction to be spontaneous, the temperature should be greater than 935 K.
$Sn{O_2}(s) + C(s)\buildrel {} \over \longrightarrow Sn(s) + C{O_2}(g)$
Standard enthalpy of reaction,
$\Delta H_{{R^\pi }}^o = (\Delta H_f^oC{O_2}(g)) - (\Delta H_f^oSn{O_2}(s))$
$\Delta H_{{R^\pi }}^o = - 394 - ( - 581) \Rightarrow 187$
Standard entropy of reaction,
$\Delta S_{{R^\pi }}^o = \Delta S_{\Pr oducts}^o - \Delta S_{{\mathop{\rm Re}\nolimits} ac\tan ts}^o$
$\Delta S_{{R^\pi }}^o = [S_{}^o(Sn(s)) + S_{}^o(C{O_2}(g))] - S_{}^o(Sn{O_2}(s)) + S_{}^o(C(s))]$
$\Delta S_{{R^\pi }}^o = [52 + 210] - [56 + 6]$
$ \Rightarrow $ 200 JK$-$1 mol$-$1
We know that, $\Delta H^\circ = T\Delta S^\circ $
$ \therefore $ $T = {{\Delta H^\circ } \over {\Delta S^\circ }}$
$ = {{187 \times 1000} \over {200}} \Rightarrow 935K$
For the reaction to be spontaneous, the temperature should be greater than 935 K.
2020
JEE Advanced
MSQ
JEE Advanced 2020 Paper 1 Offline
In thermodynamics, the p-V work done is given by
$w = - \int {dV{p_{ext}}} $
For a system undergoing a particular process, the work done is
$w = - \int {dV\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)} $
This equation is applicable to a
$w = - \int {dV{p_{ext}}} $
For a system undergoing a particular process, the work done is
$w = - \int {dV\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)} $
This equation is applicable to a
A.
system that satisfies the van der Walls' equation of state
B.
process that is reversible and isothermal
C.
process that is reversible and adiabatic
D.
process that is irreversible and at constant pressure
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
Which of the following statements regarding the first law of thermodynamics is correct?
A.
The energy of the isolated system plus the energy of the surrounding is constant.
B.
The energy of the isolated system minus the energy of the surrounding is constant.
C.
The energy of an isolated system is constant.
D.
The energy of an isolated system varies.
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
For the reactions,
$ \begin{aligned} 2 \mathrm{Cl}(g) & \longrightarrow \mathrm{Cl}_2(g) \\ \mathrm{CO}_2(g) & \longrightarrow \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \end{aligned} $
What are the signs of $\Delta S$, respectively?
A.
Positive and positive
B.
Positive and negative
C.
Negative and positive
D.
Negative and negative
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
Which of the following statement is correct?
A.
$\Delta G$ is equal to $\Delta G^{\circ}$ when the system is at the standard state.
B.
$\Delta G^{\circ}$ is zero when the system is at equilibrium.
C.
$\Delta G$ measures activation energy of a reaction.
D.
When $\Delta G$ is positive, the reaction should proceed forward to form more product.
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
What will be the $\Delta U$ value, when one mole of oxygen $\left(\mathrm{O}_2\right)$ is going from $-20^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ at constant volume? (Molar heat capacity for oxygen $\simeq 20.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ )
A.
2496 J
B.
20.8 J
C.
416 J
D.
1248 J