Thermodynamics practice · Hard

Determine $\Delta G^{\circ}$ for the following reaction

$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})$

$\Delta H^{\circ}=-282.84 \mathrm{~kJ} \text { given, } \mathrm{S}^{\circ} \mathrm{CO}_2=213.8$

$\mathrm{~S}^{\circ} \mathrm{O}_2=205.8 \mathrm{JK} / \mathrm{mol}$

Standard entropy of $X_2, Y_2$ and $X Y_4$ are 30, 60 and $110 \mathrm{~J} / \mathrm{K} \mathrm{~mol}$ respectively. For the reaction $\frac{1}{2} \mathrm{X}_2+2 \mathrm{Y}_2 \rightarrow \mathrm{XY}_4, \Delta \mathrm{H}=-40 \mathrm{~kJ}$. To be at equilibrium, the temperature will be:

Geological conditions are sometimes so extreme that quantities neglected in normal laboratory experiments take on an overriding importance. For example, consider the formation of diamond under geophysically typical conditions. The density of graphite is $2.4 \mathrm{~g} / \mathrm{cm}^3$ and that of diamond is $3.6 \, \mathrm{g} / \mathrm{cm}^3$ at a certain temperature and $500 \mathrm{~kbar}.$ By how much does $\Delta U_{\text {trans}}$ differs from $\Delta H_{\text {trans}}$ for the graphite to diamond transition?

Statement I Based on the following thermochemical
reactions;
$\mathrm{HF}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_2 \mathrm{O}(l)+\mathrm{F}^{-}(a q) ; \Delta H^{\circ}=-16.4 \mathrm{~k~cal}$

$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_2 \mathrm{O}(l), \Delta H^{\circ}=-13.7 \mathrm{~k~cal}$

Ionisation of $\mathrm{HF}(\mathrm{aq})$ is an exothermic reaction.

Statement II Enthalpy decrease in hydration of $\mathrm{H}^{+}(a q)$ and $\mathrm{F}^{-}(a q)$ exceeds the ionisation enthalpy of $\mathrm{HF}(a q)$.

The heats of combustion of carbon and carbon moxoide are – 393.5 and $- 283.5\,kJ\,mo{l^{ - 1}}$, respectively. The heat of formation (in kJ) of carbon monoxide per mole is

The reaction, $MgO(s) + C(s) \to Mg(s) + CO(g)$, for which ${\Delta _r}{H^ \circ } = + 491.1kJ\mathrm{ }mo{l^{ - 1}}$ and ${\Delta _r}{S^ \circ } = 198.0J{K^{ - 1}}mo{l^{ - 1}}$ , is not feasible at 298 K. Temperature above which reaction will be feasible is

A system undergoes a two-step process. In step 1, $15 \mathrm{~J}$ of work is done on the system as its internal energy increases by a total of $30 \mathrm{~J}.$ In step 2, at constant volume, the system decreases its internal energy by $20 \mathrm{~J}.$ Which statement is true about these changes?

The carbon dioxide exhaled in the breath of astronaut is often removed from the spacecrafts by reaction with lithium hydroxide.

$2 \mathrm{LiOH}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightarrow \mathrm{Li}_2 \mathrm{CO}_3(\mathrm{s})+\mathrm{H}_2 \mathrm{O}({l})$

Assume that each astronaut requires $2100 \mathrm{~kcal}$ of energy per day. Further assume that this energy is obtained only from the combustion of glucose into $\mathrm{CO}_2(\mathrm{g})$ and $\mathrm{H}_2 \mathrm{O}(l).$ The standard enthalpies of formation of glucose, $\mathrm{CO}_2$ and water are $-1280,-395$ and $-285 \mathrm{~kJ} / \mathrm{mol},$ respectively. The minimum mass of $\mathrm{LiOH}$ required per astronaut per day to react completely with all the $\mathrm{CO}_2$ produced is

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If ${C_V} = 28\mathrm{ }\,\,J{K^{ - 1}}\mathrm{ }mo{l^{ - 1}},$ calculate $\Delta U\mathrm{ }$ and $\Delta pV$ for this process, $(R = 8.0\mathrm{ }\,J{K^{ - 1}}\mathrm{ }mo{l^{ - 1}})$

A sample of argon gas at $1 \mathrm{~atm}$ pressure and $27^{\circ} \mathrm{C}$ expands reversibly and adiabatically from $1.25 \mathrm{~dm}^3$ to $2.50 \mathrm{~dm}^3.$ Calculate the enthalpy change in this process. $\mathrm{C}_{\mathrm{v}, \mathrm{m}}$ for argon is $12.48 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}$

Using the data provided, calculate the multiple bond energy $(kJ\,mo{l^{ - 1}})$of a C≡C bond in ${\mathrm{C}_2}{\mathrm{H}_2}.$ Given that the heat of formation of ${\mathrm{C}_2}{\mathrm{H}_2}$ is $225kJmo{l^{ - 1}}$ (take the bond energy of C─H bond as $350\,\,kJ\,\,mo{l^{ - 1}}.):$
$2\mathrm{C}\left( s \right) \to 2\mathrm{C}\left( \mathrm{g} \right)\Delta H = 1410\,\,\mathrm{kJ\,\,mo}{\mathrm{l}^{ - 1}}$
$2\mathrm{C}\left( s \right) \to 2\mathrm{C}\left( \mathrm{g} \right)\Delta H = 1410\,\,\mathrm{kJ\,\,mo}{\mathrm{l}^{ - 1}}$
${\mathrm{H}_2}\left( \mathrm{g} \right) \to 2\mathrm{H}\left( \mathrm{g} \right)\Delta H = 330\,\,\mathrm{kJ\,\,mo}{\mathrm{l}^{ - 1}}$

When one mole of monoatomic ideal gas at T temperature undergoes adiabatic change under a constant external pressure of 1 atm change in volume is from 1 L to 2 L, the final temperature in Kelvin would be

The value of $\Delta H^\circ$ for the reaction $2 A_{(g)}+B_{(g)} \rightleftarrows A_2 B_{(g)}$ for which $K_p=1.0 \times 10^{-10} \mathrm{~atm}^{-2}$ and $\Delta S^\circ=5 \text{ JK}^{-1}$ and $T=300 \text{ K}$ is

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $C{H_3}OH\left( l \right) + \frac{3}{2}{O_2}\left( g \right) \to C{O_2}\left( g \right) + 2{H_2}O\left( l \right)$ At 298 K, standard Gibb’s energies of formation for $C{H_3}OH\left( l \right),\,{H_2}O\left( l \right)\,\,and\,\,C{O_2}\left( g \right)$ are -166.2, -237.2 and $- 394.4\mathrm{ }kJ\mathrm{ }mo{l^{ - 1}}$ respectively. If standard enthalpy of combustion of methanol is $- 726kJ\mathrm{ }mo{l^{ - 1}}$ efficiency of the fuel cell will be:

$2.56 \mathrm{~g}$ of sulphur is burnt in a constant volume calorimeter with excess of $\mathrm{O}_2(g)$. Temperature increases from $21.25^{\circ} \mathrm{C}$ to $26.72^{\circ} \mathrm{C}$. The bomb has a heat capacity of $923 \mathrm{~J} \mathrm{~K}^{-1}$. Calorimeter contains $815 \mathrm{~g}$ of water. Thus, change in internal energy per mole of $\mathrm{SO}_2$ formed for the reaction is

$\mathrm{S}_8(s)+8 \mathrm{O}_2(g) \longrightarrow 8 \mathrm{SO}_2(g)$

(Specific heat of water is $4.184 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~g}^{-1}$ )

For the reaction, $2 A(g)+B(g) \longrightarrow 2 D(g) ;$
$\Delta U^{\circ}=-10.5 \mathrm{~kJ} \text { and } \Delta S^{\circ}=-44.1 \mathrm{~J~K}^{-1} .$

The $\Delta G^{\circ}$ for the reaction will be, also predict whether the reaction may occur spontaneously.

$\text {(} R=8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text { and } T=298 \mathrm{~K} \text {)}$

Enthalpy of vaporisation for water is 186.5 kJ $\mathrm{mo}{\mathrm{l}^{ - 1}}$ . The entropy change during vaorisation is …kJ ${\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}$ .

Calculate proton affinity of ${NH}_3(\mathrm{g})$ from the following data.

$\begin{aligned} & \Delta H_{\text {dissociation }} {H}_2=218 \mathrm{~kJ} \mathrm{~mole}^{-1} \\ & \Delta H_{\text {dissociation }} {Cl}_2=124 \mathrm{~kJ} \mathrm{~mole}^{-1} \\ & \Delta H_{\mathrm{f}}^\circ \text { of } {NH}_3(\mathrm{g})=-46 \mathrm{~kJ} \mathrm{~mole}^{-1} \\ & \Delta H_{\mathrm{f}}^\circ \text { of } {NH}_4 {Cl}(\mathrm{s})=-314 \mathrm{~kJ} \mathrm{~mole}^{-1} \end{aligned}$

Ionization energy of $H=1310 \mathrm{~kJ} \mathrm{~mole}^{-1}$

E.A. of ${Cl}(\mathrm{g})=-348 \mathrm{~kJ} \mathrm{~mole}^{-1}$

Lattice energy of ${NH}_4 {Cl}(\mathrm{s})=-683 \mathrm{~kJ} \mathrm{~mole}^{-1}$

For the complete combustion of ethanol, ${C_2}{H_5}OH(l) + 3{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l),$ the amount of heat produced as measured in bomb calorimeter, is $1364.47\,k{J^{ - 1}}\,mo{l^{ - 1}}$ at $25^\circ C$. Assuming ideality the enthalpy of combustion, ${\Delta _C}H,$ for the reaction will be
($R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}$)

STATEMENT-1 : The standard free energy changes of all spontaneously occuring reactions are negative.

STATEMENT-2 : The standard free energies of the elements in their standard states at 1 bar and 298 K are taken as zero.

For an isothermal free expansion of an ideal gas correct option is:

Statement I In the following thermochemical reactions,

$\begin{aligned} & C(s)+\mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \Delta H=-94 \mathrm{~k~cal} \mathrm{~mol}^{-1} \\ & \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}(g) ; \Delta H=-26 \mathrm{~k~cal} \mathrm{~mol}^{-1} \\ & \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \end{aligned}$

Enthalpy of combustion of $\mathrm{CO}(g)$ is $-68 \mathrm{~k~cal} \mathrm{~mol}^{-1}$.

Statement II $\mathrm{CO}_2(g)$ is an exothermic compound.

One mole of an ideal gas $(\gamma=1.4)$ is expanded isothermally at $27^{\circ} \mathrm{C}$ till its volume is doubled. It is then adiabatically compressed to its original volume. The magnitude of total work done by the gas is $\left(\ln 2=0.7,2^{0.4}=\frac{4}{3}, \mathrm{R}=8.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol}\right)$

An ideal gas is expanded irreversibly from $5 \mathrm{~L}$ to $10 \mathrm{~L}$ against a constant external pressure of $1 \mathrm{~bar}.$ The value of heat involved $(q)$ in this isenthalpic process is

The reaction between ${H}_2$ and ${O}_2$ to produce water vapour is given by the equation : $2 {H}_2({~g})+{O}_2({~g}) \longrightarrow 2 {H}_2 {O}({g})$
The work done at STP is :

The enthalpy of vapourisation of water = 45.953 kJ/mol. At 1 atm pressure the boiling point of water is 373 K. What is the boiling point of water when pressure equal to 1/2 atm in Kelvin in nearest possible integers? Given: $\log 2=0.3$

One mole of an ideal gas undergoes a reversible process in which the entropy of the gas changes with absolute temperature $T$ as $S=a T+C_{\mathrm{v}, \mathrm{m}}$ $\ln T,$ where $a$ is a positive constant. If $T=T_0$ at $V=V_0,$ then the volume dependence of the gas on temperature in this process is

Heat of combustion of glucose is $-2520 \mathrm{~kJ} / \mathrm{mol}$ then calculate the moles of required $\mathrm{O}_2$ when amount of heat evolved is $504 \mathrm{~kJ}$ ?

An athlete in the weight room lifts a $50 \mathrm{~kg}$ mass through a vertical distance of $2.0 \mathrm{~m}.$ The mass is allowed to fall through the $2.0 \mathrm{~m}$ distance while coupled to an electrical generator. The electrical generator produces an equal amount of electrical work, which is used to produce aluminium by Hall electrolytic process.

$\begin{aligned} & {Al}_2 {O}_3 \text {(Solution)}+3 {C}(\text {Graphite}) \rightarrow 2 {Al}({l})+ 3 {CO}(\mathrm{g}) ; \Delta G^{\circ}=600 \mathrm{~kJ} \end{aligned}$

How many times must the athlete lift the $50 \mathrm{~kg}$ mass to provide sufficient Gibbs energy to produce $27 \mathrm{~g} \, {Al}?$ $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$

Given that
(i) $\mathrm{CH}_3 \mathrm{CHO}+\frac{5}{2} \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$; $\Delta \mathrm{H}=-1168 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(ii) $\mathrm{CH}_3 \mathrm{COOH}+2 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$ ; $\Delta \mathrm{H}=-876 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\Delta \mathrm{H}$ for the reaction, $\mathrm{CH}_3 \mathrm{CHO}+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{CH}_3 \mathrm{COOH}$, is