Thermodynamics
The enthalpy change for the conversion of $\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g})$ to $\mathrm{Cl}^{-}$(aq) is ($-$) ___________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (Nearest integer)
Given : $\Delta_{\mathrm{dis}} \mathrm{H}_{\mathrm{Cl}_{2(\mathrm{~g})}^{\theta}}^{\ominus}=240 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{eg}} \mathrm{H}_{\mathrm{Cl_{(g)}}}^{\ominus}=-350 \mathrm{~kJ} \mathrm{~mol}^{-1}$,
${\mathrm{\Delta _{hyd}}H_{Cl_{(g)}^ - }^\Theta = - 380}$ $\mathrm{kJ~mol^{-1}}$
Explanation:
Given $\mathrm{C_V}=20 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Explanation:
$\mathrm{T_1=300~K}$
$\mathrm{w = 3}$ kJ/mole
$\mathrm{w=nC_v\Delta T}$
$3000=1\times20\times(300-\mathrm{T_2})$
$300-\mathrm{T_2}=150$
$\mathrm{T_2=150~K}$
When 2 litre of ideal gas expands isothermally into vacuum to a total volume of 6 litre, the change in internal energy is ____________ J. (Nearest integer)
Explanation:
For isothermal process of an ideal gas; $\mathrm{\Delta E=0}$
28.0 L of CO$_2$ is produced on complete combustion of 16.8 L gaseous mixture of ethene and methane at 25$^\circ$C and 1 atm. Heat evolved during the combustion process is ___________ kJ.
Given : $\mathrm{\Delta H_c~(CH_4)=-900~kJ~mol^{-1}}$
$\mathrm{\Delta H_c~(C_2H_4)=-1400~kJ~mol^{-1}}$
Explanation:
Let, Volume of C2H4 is x litre
$ \begin{aligned} &\Rightarrow 28=16.8+\mathrm{x} \\\\ &\mathrm{x}=11.2 \mathrm{~L} \\\\ &\mathrm{n}_{\mathrm{CH}_4}=\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{1 \times 5.6}{0.082 \times 298}=0.229 \text { mole } \\\\ &\mathrm{n}_{\mathrm{C}_2 \mathrm{H}_2}=\frac{11.2}{0.082 \times 298}=0.458 \text { mole } \\\\ &\therefore \text { Heat evolved }=0.229 \times 900+0.458 \times 1400 \\\\ &=206.1+641.2 \\\\ &=847.3 \mathrm{~kJ} \end{aligned} $
An athlete is given 100 g of glucose (C$_6$H$_{12}$O$_6$) for energy. This is equivalent to 1800kJ of energy. The 50% of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is ____________ g (Nearest integer)
Assume that there is no other way of consuming stored energy.
Given : The enthalpy of evaporation of water is 45 kJ mol$^{-1}$
Molar mass of C, H & O are 12, 1 and 16 g mol$^{-1}$.
Explanation:
Extra energy used to convert $\mathrm{H}_2 \mathrm{O(l)}$ into $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ into $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$
$ \begin{aligned} & =\frac{1800}{2}=900 \mathrm{~kJ} \\\\ & \Rightarrow 900=\mathrm{n}_{\mathrm{H}_2 \mathrm{O}} \times 45 \\\\ & \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=\frac{900}{45}=20 \text { mole } \\\\ & \mathrm{W}_{\mathrm{H}_2 \mathrm{O}}=20 \times 18=360 \mathrm{~g} \end{aligned} $
One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is __________ J (nearest integer)
Given ; $\log2=0.3$
$\ln10=2.3$
Explanation:
$2 \rightarrow 3 \Rightarrow$ Isochoric process
$3 \rightarrow 1 \Rightarrow$ Isothermal process
$\mathrm{W}=\mathrm{W}_{1 \rightarrow 2}+\mathrm{W}_{2 \rightarrow 3}+\mathrm{W}_{3 \rightarrow 1}$
$=\left(-\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)+0\left[-\mathrm{P}_1 \mathrm{~V}_1 \ln \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)\right]\right)$
$=\left[-1 \times(40-20)+0+\left[-1 \times 20 \ln \left(\frac{20}{40}\right)\right]\right]$
$=-20+20 \ln 2$
$=-20+20 \times 2.3 \times 0.3$
$=-6.2$ bar $L$
$|\mathrm{W}|=6.2$ bar $1=620 \mathrm{~J}$
Following figure shows spectrum of an ideal black body at four different temperatures. The number of correct statement/s from the following is ____________.
A. $\mathrm{T_4 > T_3 > T_2 > T_1}$
B. The black body consists of particles performing simple harmonic motion.
C. The peak of the spectrum shifts to shorter wavelength as temperature increases.
D. ${{{T_1}} \over {{v_1}}} = {{{T_2}} \over {{v_2}}} = {{{T_3}} \over {{v_3}}} \ne $ constant
E. The given spectrum could be explained using quantisation of energy.
Explanation:
B. It is incorrect as particles do not undergo simple harmonic motion.
C. It is correct
D. It is incorrect
E. It is correct
For independent process at 300 K
| Process | $\mathrm{\Delta H/kJ~mol^{-1}}$ | $\mathrm{\Delta S/J~K^{-1}}$ |
|---|---|---|
| A | $-25$ | $-80$ |
| B | $-22$ | 40 |
| C | 25 | $-50$ |
| D | 22 | 20 |
The number of non-spontaneous process from the following is __________
Explanation:
Processes $\mathrm{C}$ and $\mathrm{D}$ are non-spontaneous.
[Use : $\ln 2=0.69$
Given : $S_\beta-S_\alpha=0$ at $0 \mathrm{~K}$ ]
Explanation:
$ \begin{aligned} & \mathrm{S}_{\alpha(600)}^{\mathrm{o}}=\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+\mathrm{C}_{\mathrm{P}(\alpha)} \ln \frac{600}{300} \\\\ & \mathrm{~S}_{\beta(600)}^{\mathrm{o}}=\mathrm{S}_{\beta(300)}^{\mathrm{o}}+\mathrm{C}_{\mathrm{P}(\beta)} \ell \mathrm{n} \frac{600}{300} \\\\ & \mathrm{~S}_{\beta(600)}^{\mathrm{o}}-\mathrm{S}_{\alpha(600)}^{\mathrm{o}}=\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+\left(\mathrm{C}_{\mathrm{P}(\beta)}-\mathrm{C}_{\mathrm{P}(\alpha)}\right) \ell \mathrm{n} 2 \\\\ & 6-5=\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+1 \times \ell \mathrm{n} 2 \\\\ & 1=\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+0.69 \end{aligned} $
So $\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}=0.31$
Explanation:
So at $600 \mathrm{~K} \quad \Delta \mathrm{G}_{\mathrm{rxn}}^{\circ}=0$ So $\Delta \mathrm{H}^{\circ}{ }_{\text {reaction (600) }}=\mathrm{T} \Delta \mathrm{S}^{\circ}{ }_{\text {reaction (600) }}$
$\Delta \mathrm{H}^{\circ}{ }_{(600)}=600 \times 1=600 \mathrm{Joule} / \mathrm{mole}$
So $\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=\Delta \mathrm{C}_{\mathrm{P}}\left(\mathrm{T}_2-\mathrm{T}_1\right)$
$\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=1 \times 300$
$\Delta \mathrm{H}_{300}=\Delta \mathrm{H}_{600}-300=600-300=300$ Joule $/$ mole.
$\mathrm{A} \rightarrow \mathrm{B}$ is an adiabatic process. If the total heat absorbed in the entire process $(\mathrm{A} \rightarrow \mathrm{B}$ and $\mathrm{B} \rightarrow \mathrm{C})$ is $\mathrm{R} T_2 \ln 10$, the value of $2 \log V_3$ is _______.
[Use, molar heat capacity of the gas at constant pressure, $C_{\mathrm{p}, \mathrm{m}}=\frac{5}{2} \mathrm{R}$ ]
Explanation:
$ \begin{aligned} &\Rightarrow \left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right) =\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{\mathrm{V}-1} \\\\ &\Rightarrow\mathrm{~T}_1 \mathrm{~V}_1^{\gamma-1} =\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\\\ &\Rightarrow600(10)^{2 / 3} =60\left(\mathrm{~V}_2\right)^{2 / 3} \\\\ &\Rightarrow\left(\mathrm{~V}_2\right)^{2 / 3} =(10)^{5 / 3} \\\\ &\Rightarrow\mathrm{~V}_2 =(10)^{5 / 2} \\\\ &\Rightarrow\mathrm{Q}_{\mathrm{AB}} =0 \end{aligned} $
$\begin{aligned} \mathrm{Q}_{\mathrm{AC}} & =\mathrm{nRT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) \\\\ & =\mathrm{RT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) .........(i)\end{aligned}$
$\begin{gathered}\text { Total heat absorbed }=\mathrm{RT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) \\\\ =\mathrm{RT}_2 \ln (10) ........(ii)\end{gathered}$
Equating equation (i) and equation (ii)
$ \begin{aligned} &\mathrm{RT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) =\mathrm{RT}_2 \ln (10) \\\\ &\Rightarrow\ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) =\ln (10) \end{aligned} $
$ \mathrm{V}_3=10 \mathrm{~V}_2 $
Substitute value of $\mathrm{V}_2$
$ \begin{aligned} & =10(10)^{5 / 2}=(10)^{7 / 2} \\\\ V_3 & =(10)^{7 / 2} \end{aligned} $
Taking $\log$ on both side, we get
$ \begin{aligned} &\Rightarrow \log \left(\mathrm{V}_3\right) =\log (10)^{7 / 2} \\\\ &\Rightarrow \log \left(\mathrm{V}_3\right) =\frac{7}{2} \log (10) \\\\ &\Rightarrow 2 \log \left(\mathrm{V}_3\right) =7 \log (10) \\\\ \therefore 2 \log \left(\mathrm{V}_3\right) =7 \end{aligned} $
If $\mathrm{T}_1=2 \mathrm{~T}_2$ and $\left(\Delta \mathrm{G}_2^{\Theta}-\Delta \mathrm{G}_1^{\Theta}\right)=\mathrm{RT}_2 \ln \mathrm{x}$, then the value of $\mathrm{x}$ is _______.
$\left[\Delta \mathrm{G}_1^{\Theta}\right.$ and $\Delta \mathrm{G}_2^{\Theta}$ are standard Gibb's free energy change for the reaction at temperatures $\mathrm{T}_1$ and $\mathrm{T}_2$, respectively.]
Explanation:
$\Rightarrow$ At $\mathrm{T}_1 \mathrm{~K}: \mathrm{K}_{\mathrm{P}_1}=\frac{4}{2}=2$
$ \begin{array}{ll} \text { At } \mathrm{T}_2 \mathrm{~K}: & \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g}) \\\\ \mathrm{t}=0 & 6 \\\\ \mathrm{t}=\infty & 6-\mathrm{y} \quad \mathrm{y}=2 \text { (from plot) } \end{array} $
$ \Rightarrow \text { At } \mathrm{T}_2 \mathrm{~K}: \mathrm{K}_{\mathrm{P}_2}=\frac{2}{4}=\frac{1}{2} $
Since
$ \begin{aligned} \Delta \mathrm{G}_1^0 & =-\mathrm{RT}_1 \ln \mathrm{KT}_1 ............(i) \\\\ \Delta \mathrm{G}_2^0 & =-\mathrm{RT}_2 \ln \mathrm{KT}_2 ............(ii) \end{aligned} $
$\begin{aligned} \Delta \mathrm{G}_2^0-\Delta \mathrm{G}_1^0 & =-\mathrm{RT}_2 \ln \mathrm{KT}_2+\mathrm{RT}_1 \ln \mathrm{KT}_1 \\\\ & =-\mathrm{RT}_2 \ln \frac{1}{2}+\mathrm{RT}_1 \ln 2 \\\\ & =-\mathrm{RT}_2 \ln \frac{1}{2}+\mathrm{R}\left(2 \mathrm{~T}_2\right) \ln 2 \\\\ & =\mathrm{RT}_2 \ln 2+2 \mathrm{RT}_2 \ln 2 \\\\ & =\mathrm{RT}_2 \ln 2+2 \mathrm{RT}_2 \ln 2\end{aligned}$
$\begin{aligned} \Delta \mathrm{G}_2^0-\Delta \mathrm{G}_1^0 & =3 \mathrm{RT}_2 \ln 2 \\\\ & =\mathrm{RT}_2\left(\ln 2^3\right) \\\\ & =\mathrm{RT}_2 \ln 8\end{aligned}$
At $T(\mathrm{~K}) 2$ mole of an ideal gas is allowed to expand reversibly and isothermally from a pressure of 10 atmospheres to 1 atmosphere. The work done (in kJ ) is $\left(R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$
$-3.82 \times 10^{-1} \times T$
$-4.82 \times 10^{-1} \times T$
$-2.82 \times 10^{-2} \times T$
$-3.82 \times 10^{-2} \times T$
For the reaction at $25^{\circ} \mathrm{C}, X_2 \mathrm{O}_4(l) \longrightarrow 2 \mathrm{XO}_2(g), \Delta U$ and $\Delta S$ are 2.1 k cal and $20 \mathrm{cal} / \mathrm{K}$ respectively. What is $\Delta \mathrm{G}$ for the reaction at the same temperature? ( $R=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ )
-2.67 k cal
+2.67 k cal
-1.67 k cal
+3.67 k cal
In. Observe the following properties.
- Molar volume
II. Mass
III. Internal energy
IV. Volume
v. Enthalpy
VI. Temperature
VII. Density
The intensive properties in the above list are
I, VI, VII only
I, IV, VI, VII only
I, III, IV, V only
II, III, V only
Enthalpy of formation of $\mathrm{CO}(g), \mathrm{CO}_2(g)$ are -110 , $-393 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The enthalpy of combustion of CO (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is
-283.0
-110.5
504
-221.2
The enthalpies of formation of gaseous $\mathrm{N}_2 \mathrm{O}$ and NO at 298 K are 82.0 and $90.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The enthalpy change of the reaction
$\mathrm{N}_2 \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow 2 \mathrm{NO}(g)$ is
The temperature of $K$ at which $\Delta G=0$, for a given reaction with $\Delta H=-20.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S=-50.0 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}$ is
A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If $\Delta H$ is the enthalpy change and $\Delta E$ is the change in internal energy, then
The value of enthalpy change $(\Delta H)$ for the reaction $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+ 3 \mathrm{H}_2 \mathrm{O}(l)$, at $27^{\circ} \mathrm{C}$ is $-1366 \cdot 5 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The value of internal energy change for the above reaction at this temperature will be
Given below are two statements: One is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$
Assertion A : The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.
Reason $\mathbf{R}$ : The value of $\Delta G ^\Theta$ becomes more on negative side as entropy is higher in liquid state than solid state.
In the light of the above statements, choose the most appropriate answer from the options given below
Which of the following relation is not correct?
$\Delta$G$^\circ$ vs T plot for the formation of MgO, involving reaction
2Mg + O2 $\to$ 2MgO, will look like :
Match List-I with List-II.
| List - I | List - II | ||
|---|---|---|---|
| (A) | Spontaneous process | (I) | $\Delta H < 0$ |
| (B) | Process with $\Delta P = 0$, $\Delta T = 0$ | (II) | $\Delta {G_{T,P}} < 0$ |
| (C) | $\Delta {H_{reaction}}$ | (III) | Isothermal and isobaric process |
| (D) | Exothermic Process | (IV) | [Bond energies of molecules in reactants] $ - $ [Bond energies of product molecules |
Choose the correct answer from the options given below :
At 25$^\circ$C and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are $-$ 3268 kJ mol$-$1 and $-$1300 kJ mol$-$1, respectively. The change in enthalpy for the reaction 3 C2H2(g) $\to$ C6H6 (I), is :
At 25$^\circ$C and 1 atm pressure, the enthalpies of combustion are as given below :
| Substance | ${H_2}$ | C (graphite) | ${C_2}{H_6}(g)$ |
|---|---|---|---|
| ${{{\Delta _c}{H^\Theta }} \over {kJ\,mo{l^{ - 1}}}}$ | $ - 286.0$ | $ - 394.0$ | $ - 1560.0$ |
The enthalpy of formation of ethane is
When 600 mL of 0.2 M HNO3 is mixed with $400 \mathrm{~mL}$ of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ___________ $\times 10^{-2}{ }\,^{\circ} \mathrm{C}$.
(Enthalpy of neutralisation $=57 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and Specific heat of water $=4.2 \,\mathrm{JK}^{-1} \mathrm{~g}^{-1}$) (Neglect heat capacity of flask)
Explanation:
600 mL × 0.2 M = 120 m mol
NaOH
400 mL × 0.1 M = 40 m mol
Heat liberated from reaction
$ =40 \times 10^{-3} \times 57 \times 10^{3} \mathrm{~J} $
Heat gained by solution $=m C \Delta T$
$ \begin{aligned} \mathrm{m}=\text { mass of solution }=\mathrm{V} \times \mathrm{d} &=1000 \times 1 \\\\ &=1000 \mathrm{~g} \end{aligned} $
Heat gained by solution $=1000 \times 4.2 \times \Delta \mathrm{T} \ldots(2)$
From (1) and (2)
Heat liberated $=$ Heat gained
$40 \times 10^{-3} \times 57 \times 10^{3}=1000 \times 4.2 \times \Delta T$
$\Delta T=54 \times 10^{-2}{ }^{\circ} \mathrm{C}$
(Rounded off to the nearest integer)
Among the following the number of state variables is ______________.
Internal energy (U)
Volume (V)
Heat (q)
Enthalpy (H)
Explanation:
A gas (Molar mass = 280 $\mathrm{~g} \mathrm{~mol}^{-1}$) was burnt in excess $\mathrm{O}_{2}$ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from $298.0 \mathrm{~K}$ to $298.45$ $\mathrm{K}$. If the heat capacity of calorimeter is $2.5 \mathrm{~kJ} \mathrm{~K}^{-1}$ and enthalpy of combustion of gas is $9 \mathrm{~kJ} \mathrm{~mol}^{-1}$ then amount of gas burnt is _____________ g. (Nearest Integer)
Explanation:
$ \begin{aligned} &=2.5 \times 10^{3} \times 0.45 \\\\ &=1.125 \mathrm{~kJ} \end{aligned} $
Considering $\Delta \mathrm{H} \simeq \Delta \mathrm{U}$
$ \Delta \mathrm{H}=9 \mathrm{~kJ} / \mathrm{mol} \simeq \Delta \mathrm{U} $
$\therefore$ Mass of gas burnt $=\frac{1.125}{9} \times 280=35 \mathrm{~g}$
The molar heat capacity for an ideal gas at constant pressure is $20.785 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$. The change in internal energy is $5000 \mathrm{~J}$ upon heating it from $300 \mathrm{~K}$ to $500 \mathrm{~K}$. The number of moles of the gas at constant volume is ____________. [Nearest integer] (Given: $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$)
Explanation:
$ \begin{aligned} &\text { and } \Delta \mathrm{U}=\mathrm{nC} \mathrm{v} \Delta \mathrm{T} \\\\ &\therefore \quad \mathrm{nC}_{\mathrm{v}}=\frac{5000}{200}=25 \end{aligned} $
and we know that
$ \begin{aligned} &C_{p}-C_{v}=R \\\\ &20.785-\frac{25}{n}=8.314 \\\\ &n=\frac{25}{(20.785-8.314)}=2 \end{aligned} $
For the reaction
$\mathrm{H}_{2} \mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})$
$\Delta U=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$.
The enthalpy change for the above reaction is ($-$) __________ $\mathrm{kJ} \,\mathrm{mol}^{-1}$ [nearest integer]
Given: $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.
Explanation:
$\Delta \mathrm{U}=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$
$ \begin{aligned} \Delta \mathrm{H} &=\Delta \mathrm{U}+\Delta \mathrm{n}_{g} \mathrm{RT} \\ &=-59.6+\frac{1 \times 8.314 \times 300}{1000} \\ &=-57.10 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
$2.4 \mathrm{~g}$ 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)
Explanation:
$\mathrm{n}_{\text {coal }}=\frac{2.4}{12}$
$\mathrm{Q}=\frac{-20(300-298)}{0.2}$
$Q=-200 \mathrm{~kJ} / \mathrm{mol}$
$x=200$
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)
Explanation:
$ \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} $
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)
Explanation:
$\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{I}), \Delta \mathrm{H}_{1}=-2220 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{C}($ 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} $
$\left|\Delta \mathrm{H}_{\mathrm{f}}\right| \simeq 104 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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)
Explanation:
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)
Explanation:
$ \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} $
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.
Explanation:
So, required $\Delta \mathrm{H}=5 \times 23.4$
$ =117 \mathrm{~kJ} $
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)
Explanation:
$ \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 $600 \mathrm{~kJ}$.
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.
Explanation:
$ \because P_{\text {ext }}=0 \quad \text { (vacuum) } $
$\therefore w=0, \quad\Delta U=0$ (as the process is isothermal)
So, $q=0$
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$-$1 mol$-$1 and log 2 = 0.3010)
Explanation:
A fish swimming in water body when taken out from the water body is covered with a film of water of weight 36 g. When it is subjected to cooking at 100$^\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]
Explanation:
$ \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} $
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)
Explanation:
$ \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} $
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)
Explanation:
$ \begin{aligned} &\Rightarrow-165 \times 10^3-\mathrm{T} \times(-505)=0 \\\\ &\Rightarrow \mathrm{T}=300 \mathrm{~K} \end{aligned} $
$2 \mathrm{~mol} \,\mathrm{of}\, \mathrm{Hg}(\mathrm{g})$ is combusted in a fixed volume bomb calorimeter with excess of $\mathrm{O}_{2}$ at $298 \mathrm{~K}$ and 1 atm into $\mathrm{HgO}(s)$. During the reaction, temperature increases from $298.0 \mathrm{~K}$ to $312.8 \mathrm{~K}$. If heat capacity of the bomb calorimeter and enthalpy of formation of $\mathrm{Hg}(g)$ are $20.00 \mathrm{~kJ} \mathrm{~K}^{-1}$ and $61.32 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$ at $298 \mathrm{~K}$, respectively, the calculated standard molar enthalpy of formation of $\mathrm{HgO}(s)$ at 298 $\mathrm{K}$ is $\mathrm{X}\, \mathrm{kJ}\, \mathrm{mol}^{-1}$. The value of $|\mathrm{X}|$ is _________ .
[Given: Gas constant $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]
Explanation:
$2Hg(g)+O_2(g) \longrightarrow 2HgO(s)$
Heat capacity of calorimeter = 20 kJ K$-1$
Rise in temperature = 14.8 K
$\Delta H^\circ=\Delta U^\circ+\Delta n_g RT$
$=-296-3\times8.3\times298\times10^{-3}$
$\simeq -303.42$ kJ
$\Delta H^\circ = \Delta H{^\circ _f}(HgO(s)) - \Delta H{^\circ _f}(Hg(g))$
$ - 303 - 42 = \Delta H{^\circ _f}(HgO(s)) - 2 \times 61.32$
$\Delta H{^\circ _f}(HgO(s)) = - 303.42 - 122.64$
$ = - 180.78$ kJ
$\left| {\Delta H{^\circ _f}(HgO(s))} \right| = 90.39$ kJ mol$-$1
$[\mathrm{R}=$ gas constant, $\mathrm{F}=$ Faraday constant, $\mathrm{T}=$ Temperature $]$
The bond enthalpies of heavy hydrogen, $\mathrm{O}-\mathrm{O}$ and $\mathrm{D}-\mathrm{O}$ are $+400,+498$ and $+490 \mathrm{kJmol}^{-1}$, respectively. The $\Delta_r H^{\circ}$ of the reaction to produce $\mathrm{D}_2 \mathrm{O}$ is
$-300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$-331 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$29.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$2.91 \mathrm{~kJ} \mathrm{~mol}^{-1}$
In the reaction, $\mathrm{H}_2 \mathrm{O}(l) E \longrightarrow \mathrm{H}_2 \mathrm{O}(s)$ at $0^{\circ} \mathrm{C}$ and 1 atom, the internal energy change is $-41 \mathrm{~kJ} / \mathrm{mol}$. What will be the value of molar enthalpy change?
$-41 \mathrm{~kJ} / \mathrm{mol}$
$41 \mathrm{~kJ} / \mathrm{mol}$
$30 \mathrm{~kJ} / \mathrm{mol}$
$-30 \mathrm{~kJ} / \mathrm{mol}$
The correct order of " $\Delta H_f^{\circ}$ " values of diamond (I), graphite (II) and fullerene (III) is
I $>$ II $>$ III
II $>$ I $>$ III
III $>$ I $>$ II
III $>$ II $>$ I
An air bag on adiabatic expansion undergoes $5 \%$ increase in its volume. The percentage change in pressure is $\left[\gamma_{\text {air }}=1.4\right]$
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