Thermodynamics practice · Mixed difficulty

In a process, $701 \mathrm{~J}$ of heat is absorbed by a system and $394 \mathrm{~J}$ of work is done by the system. What is the change in internal energy for the process?

Based on following thermochemical reaction:

$\begin{aligned} & {H}_2 {O}(\mathrm{g})+{C}(\mathrm{s}) \rightarrow {CO}(\mathrm{g})+{H}_2(\mathrm{g}) ; \Delta {H}=131 \mathrm{~KJ} \\ & {CO}(\mathrm{g})+\frac{1}{2} {O}_2(\mathrm{g}) \rightarrow {CO}_2(\mathrm{g}) ; \Delta {H}=-282 \mathrm{~KJ} \\ & {H}_2(\mathrm{g})+\frac{1}{2} {O}_2(\mathrm{g}) \rightarrow {H}_2 {O}(\mathrm{g}) ; \Delta {H}=-242 \mathrm{~KJ} \\ & {C}(\mathrm{s})+{O}_2(\mathrm{g}) \rightarrow {CO}_2(\mathrm{g}) ; \Delta {H}=x \mathrm{~KJ} \end{aligned}$

The value of $x$ will be

The standard molar entropy of an ideal gas $\left(\gamma=\frac{4}{3}\right)$ is $2.5 \mathrm{~cal} / \mathrm{K}-\mathrm{mol}$ at $25^{\circ} \mathrm{C}$ and $1 \mathrm{~bar}.$ The standard molar entropy of the gas at $323^{\circ} \mathrm{C}$ and 1 bar is $(\ln 2=0.7)$

${C}+\frac{1}{2} {O}_2 \rightarrow {CO} ; \Delta {H}=-42 \mathrm{~kJ}$

${CO}+\frac{1}{2} {O}_2 \rightarrow {CO}_2 ; \Delta {H}=-24 \mathrm{~kJ}$

The heat of formation of ${CO}_2$ is

Calculate the lattice energy forthe reaction,

$\mathrm{Li}^{+}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g}) \longrightarrow \mathrm{LiCl}(\mathrm{s})$

From the following data:

$\Delta_{\mathrm{s}} \mathrm{H}^{\circ}(\mathrm{Li})=160.67 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \quad \frac{1}{2} \Delta_{\mathrm{a}} \mathrm{H}^{\circ}\left(\mathrm{Cl}_2\right)=122.17 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{\mathrm{i}} \mathrm{H}^{\circ}(\mathrm{Li})=520.07 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ; $\quad$ $\Delta_{\mathrm{eg}} \mathrm{H}^{\circ}(\mathrm{Cl})=-365.26 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\text { and } \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{LiCl})=-401.66 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The heat change taking place during the reaction $\;{\mathrm{H}_2}\mathrm{O}\left( l \right)$$\to {\mathrm{H}_2}\mathrm{O(g)}$ is: [Given,$\Delta {H_f}$ of ${\mathrm{H}_2}\mathrm{O}(g)$ = -57 kcal, $\Delta {H_f} = {\mathrm{H}_2}\mathrm{O}\left( l \right)$ = -68.3 kcal]

0.75 mole of an ideal gas expands at $27^{\circ} \mathrm{C}$ from $15 \mathrm{~L}$ to $25 \mathrm{~L}$, then maximum work done is

Calculate the standard enthalpy of solution of $\mathrm{AgCl}(\mathrm{s})$ in water $\Delta \mathrm{H}^\circ_{\mathrm{f}}(\mathrm{AgCl}, \mathrm{s})=-127.07 \mathrm{~kJ} \mathrm{~mol}^{-1}$, $\Delta \mathrm{H^\circ_f}\left(\mathrm{Ag}^{+}, \mathrm{aq}\right)=105.58 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{H}^\circ_\mathrm{f}\left(\mathrm{Cl}^{-}, \mathrm{aq}\right)=-167.35 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

A sample of $100 {~g} {H}_2 \mathrm{O}$ is slowly heated from $27^{\circ} {C}$ to $87^{\circ} {C}$. Calculate the change in entropy during heating (Specific heat of water $=4200 {~J} / {kg}-{K}$ ).

Assertion (A): The difference between $\Delta H$ and $\Delta U$ is not significant for systems consisting of only solids or liquids.
Reason (R): Solids and liquids do not suffer any significant volume changes upon heating.

Assertion: When ideal gas expand from $P_1, V_1, T_1$ to $P_2, V_2, T_2$ in multiple steps, work done is high where number of steps are high.
Reason: Work is path function.

The data below refers to gas phase reaction at constant pressure at $25^{\circ} \mathrm{C}.$

$\begin{aligned} & \text{(a) } \mathrm{CH}_3-\mathrm{CH}_3 \rightarrow \mathrm{CH}_3-\mathrm{CH}_2+\mathrm{H}; \Delta H_1=+420 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \text{(b) } \mathrm{CH}_3-\mathrm{CH}_2 \rightarrow \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}; \Delta H_2=+168 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

From these data, the enthalpy change $\Delta H$ for the reaction $2 \mathrm{CH}_3-\mathrm{CH}_2 \rightarrow \mathrm{CH}_3-\mathrm{CH}_3+\mathrm{CH}_2=\mathrm{CH}_2$ is

For a reaction ${Y}_2{O}_4(\ell) \longrightarrow 2 {YO}_2(\mathrm{g})$

$\Delta {E}=2.1 \mathrm{~KCal}, \Delta {S}=20 \mathrm{~calK}^{-1}$ at $300 \mathrm{~K}, \Delta {G}$ will be:

The enthalpies of formation of $\mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})$ and $\mathrm{C}_6 \mathrm{H}_6(\mathrm{~g})$ at $298 \mathrm{~K}$ are 230 and $85 \mathrm{~kJ} / \mathrm{mol}$ respectively. The enthalpy change for the reaction is

The enthalpies of combustion of formaldehyde and paraformaldehyde (a polymer of formaldehyde) are $-134$ and $-732 \mathrm{~kcal} / \mathrm{mol},$ respectively. The enthalpy of polymerization per mole of paraformaldehyde is $-72 \mathrm{~kcal}.$ The molecular formula of paraformaldehyde is

Assertion :- Entropy increases with increase in temperature of gas.

Reason :- If $q_{\text{rev}}$ is same, then entropy of that system is higher for which temperature is
higher.

A system $\mathrm{X}$ undergoes following changes

$\underset{\left(\mathrm{P}_1 \mathrm{V}_1 \mathrm{T}_1\right)}{\mathrm{X}} \longrightarrow \underset{\left(\mathrm{P}_2 \mathrm{V}_2 \mathrm{T}_1\right)}{\mathrm{W}} \longrightarrow \underset{\left(\mathrm{P}_3 \mathrm{V}_2 \mathrm{T}_2\right)}{\mathrm{Z}} \longrightarrow \underset{\left(\mathrm{P}_1 \mathrm{V}_1 \mathrm{T}_1\right)}{\mathrm{X}}$

The overall process may be called as

Assertion :- Entropy is always constant for a closed system.

Reason :- Closed system is always reversible.

1 mole of an ideal gas undergoes reversible isothermal expansion from an initial volume $V_1$ to a final volume $10 V_1$ and does $10 {~kJ}$ of work. The initial pressure was $1 \times 10^7 {~Pa}$. Calculate $V_1$.

Assertion :- Adiabatic free expansion of an ideal gas is irreversible.

Reason :- PV = constant for reversible adiabatic expansion.

One mole of an ideal gas expands isothermally and reversible from 2 lit. to 20 lit. at $300 \mathrm{~K}$. If the final pressure of the gas is I bar, the work done by the gas is

The enthalpy of atomization of graphite is $698.6 \, \mathrm{kJ} / \mathrm{mol}$ and the mean bond enthalpy of $\mathrm{C}-\mathrm{C}$ bond in diamond is $348.4 \mathrm{~kJ} / \mathrm{mol}.$ The enthalpy of conversion of graphite into diamond is

1 mole of gas occupying 3 L volume is expanded against a constant external pressure of 1 atm to a volume of 15 litre. The work done by the system is:

$\mathrm{H}_2$ gas is used as a fuel. $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{O}_2$ are mixed in stoichiometric proportion and react completely to form $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ releasing $58 \mathrm{~k~cal}$ of heat in a constant pressure calorimeter at $298 \mathrm{~K}$. Thus, temperature of hydrogen-oxygen flame is

$\Delta_r H^{\circ}$ of $\mathrm{CS}_2$ is $117.36 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{\mathrm{f}} \mathrm{H}^{\circ}$ of $\mathrm{C}(\mathrm{g})$ is $716.682 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_f H^{\circ}$ of $\mathrm{S}(g)$ is $278.805 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Thus, $(\mathrm{C}=\mathrm{S})$ bond enthalpy in $\mathrm{CS}_2$ is

A reversible heat engine absorbs $40 \mathrm{~kJ}$ of heat at $500 \mathrm{~K}$ and performs $10 \mathrm{~kJ}$ of work rejecting the remaining amount to the sink at $300 \mathrm{~K}.$ The entropy change for the universe is

Which represents the largest amount of energy?

Bond energies of $(\mathrm{H}-\mathrm{H}),(\mathrm{O}=\mathrm{O})$ and $(\mathrm{O}-\mathrm{H})$ are 105,120 and $220 \mathrm{kcal} / \mathrm{mol}$, respectively then $\Delta \mathrm{H}$ of the reaction, $2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$;

The enthalpy of formation of ${HCl}(\mathrm{g})$ from the following reaction ${H}_2(\mathrm{g})+{Cl}_2(\mathrm{g}) \rightarrow 2 {HCl}(\mathrm{g})+44 \mathrm{~kcal}$ is

A: All reactions which are exothermic are spontaneous.
R: All reactions in which entropy increases are spontaneous.