Mix Examples-Thermodynamics and Thermochemistry Questions in English

(C) The difference in enthalpy of combustion for each successive $CH_2$ group is calculated as follows:
$\Delta H_c(C_2H_6) - \Delta H_c(CH_4) = -368.4 - (-210.8) = -157.6 \ k \ cal \ mol^{-1}$
$\Delta H_c(C_3H_8) - \Delta H_c(C_2H_6) = -526.2 - (-368.4) = -157.8 \ k \ cal \ mol^{-1}$
Average value for $\Delta H_c(-CH_2-) = \frac{-157.6 + (-157.8)}{2} = -157.7 \ k \ cal \ mol^{-1}$
For hexane $(C_6H_{14})$,we add three $-CH_2-$ groups to propane $(C_3H_8)$:
$\Delta H_c(C_6H_{14}) = \Delta H_c(C_3H_8) + 3 \times \Delta H_c(-CH_2-)$
$\Delta H_c(C_6H_{14}) = -526.2 + 3(-157.7) = -526.2 - 473.1 = -999.3 \ k \ cal \ mol^{-1}$
Rounding to the nearest integer,the value is $-1000 \ k \ cal \ mol^{-1}$.

(B) For a combustion reaction,the enthalpy change $\Delta H$ is negative (exothermic).
The change in the number of moles of gas is $\Delta n_g = (16 + 18) - (2 + 25) = 34 - 27 = +7$.
Since $\Delta n_g > 0$,the entropy change $\Delta S$ is positive $(+ve)$.
As the reaction is spontaneous in an automobile engine,the Gibbs free energy change $\Delta G$ is negative $(-ve)$.

(D) The vaporization of water is a phase change process occurring at constant temperature $(373 \ K)$.
For an ideal gas,internal energy $(E)$ is a function of temperature only,so $\Delta E = 0$ for isothermal processes.
However,for real substances like water undergoing a phase change,the internal energy changes because the intermolecular forces are overcome during vaporization.
Therefore,the Assertion is incorrect because $\Delta E \neq 0$ for the vaporization of water.
The Reason is also incorrect because $\Delta E = 0$ only strictly applies to ideal gases in isothermal processes,and even then,it does not account for phase changes where potential energy changes.

(B) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta E + \Delta n_g RT$.
For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - (1 + 1) = 0$.
Since $\Delta n_g = 0$,we have $\Delta H = \Delta E$.
Thus,the Assertion is correct.
The Reason states that all reactants and products are gases,which is true,but this alone does not guarantee $\Delta H = \Delta E$. The equality holds only because $\Delta n_g = 0$. Therefore,the Reason is not the correct explanation for the Assertion.

(D) In an adiabatic expansion,the system does work at the expense of its internal energy,leading to a decrease in temperature. Thus,$T_C < T_A$.
For an adiabatic process:
$q = 0$,$\Delta U = W$
Since it is an expansion,$W < 0$,therefore $\Delta U < 0$.
Since $\Delta U = nC_{vm} \Delta T$,we have $nC_{vm} \Delta T < 0$,which implies $\Delta T < 0$.
Therefore,$T_C - T_A < 0$,or $T_C < T_A$.
Option $D$ states $T_C > T_A$,which is incorrect.

(B) The reaction is $A_{(\ell)} \longrightarrow 2 B_{(g)}$.
Given: $\Delta U = 2.1 \; kcal \; mol^{-1}$,$\Delta S = 20 \; cal \; K^{-1} \; mol^{-1} = 0.02 \; kcal \; K^{-1} \; mol^{-1}$,$T = 300 \; K$,$R = 2 \; cal \; K^{-1} \; mol^{-1} = 0.002 \; kcal \; K^{-1} \; mol^{-1}$.
Change in number of moles of gas,$\Delta n_g = 2 - 0 = 2$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$:
$\Delta H = 2.1 + (2 \times 0.002 \times 300) = 2.1 + 1.2 = 3.3 \; kcal \; mol^{-1}$.
Now,using the Gibbs free energy equation $\Delta G = \Delta H - T \Delta S$:
$\Delta G = 3.3 - (300 \times 0.02) = 3.3 - 6.0 = -2.7 \; kcal \; mol^{-1}$.

(N/A) The process is $H_2O_{(l)} \rightarrow H_2O_{(g)}$.
The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging for internal energy change: $\Delta U = \Delta H - \Delta n_g RT$.
Here,$\Delta H = 41 \ kJ \ mol^{-1}$,$R = 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$,$T = 373 \ K$,and $\Delta n_g = 1 - 0 = 1$.
Substituting the values:
$\Delta U = 41 \ kJ \ mol^{-1} - (1 \times 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1} \times 373 \ K)$
$\Delta U = 41 \ kJ \ mol^{-1} - 3.101 \ kJ \ mol^{-1}$
$\Delta U = 37.899 \ kJ \ mol^{-1} \approx 37.9 \ kJ \ mol^{-1}$.

(B) The heat absorbed by the calorimeter is given by $q = C_v \times \Delta T$.
Given $C_v = 20.7 \ kJ \ K^{-1}$ and $\Delta T = (299 - 298) \ K = 1 \ K$.
Heat released by the reaction $(q_{rxn})$ is equal to $-q_{calorimeter} = -20.7 \ kJ \ K^{-1} \times 1 \ K = -20.7 \ kJ$.
This is the internal energy change $(\Delta U)$ for the combustion of $1 \ g$ of graphite.
For $1 \ mol$ of graphite $(12 \ g)$,the internal energy change $(\Delta U)$ is:
$\Delta U = \frac{-20.7 \ kJ}{1 \ g} \times 12 \ g \ mol^{-1} = -248.4 \ kJ \ mol^{-1}$.
Since the reaction is $C \ (graphite) + O_{2(g)} \rightarrow CO_{2(g)}$,the change in moles of gaseous species is $\Delta n_g = 1 - 1 = 0$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$,since $\Delta n_g = 0$,we have $\Delta H = \Delta U$.
Therefore,$\Delta H = -248.4 \ kJ \ mol^{-1}$.

(N/A) The process of evaporation is represented as: $H_2O_{(l)} \rightarrow H_2O_{(g)}$.
Number of moles in $18 \ g$ of $H_2O$ is $n = \frac{18 \ g}{18 \ g \ mol^{-1}} = 1 \ mol$.
Heat supplied to evaporate $18 \ g$ of water at $298 \ K$ is $q = n \times \Delta_{vap} H^{\ominus} = 1 \ mol \times 44.01 \ kJ \ mol^{-1} = 44.01 \ kJ$.
To calculate the internal energy of vaporisation $(\Delta_{vap} U)$,we use the relation: $\Delta_{vap} U = \Delta_{vap} H^{\ominus} - \Delta n_g RT$.
Here,$\Delta n_g = 1$ (since $1 \ mol$ of gas is produced from $1 \ mol$ of liquid).
$\Delta_{vap} U = 44.01 \ kJ - (1 \ mol) \times (8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) \times (298 \ K)$.
$\Delta_{vap} U = 44.01 \ kJ - 2.48 \ kJ = 41.53 \ kJ$.

The change takes place as follows:
Step-$1$: $1 \ mol \ H_2O(l, 100^{\circ} C) \rightarrow 1 \ mol \ H_2O(l, 0^{\circ} C)$,Enthalpy change $\Delta H_1 = n \times C_p \times \Delta T = 1 \ mol \times 18 \ g \ mol^{-1} \times 4.2 \ J \ g^{-1} {\circ} C^{-1} \times (0^{\circ} C - 100^{\circ} C) = -7560 \ J \ mol^{-1} = -7.56 \ kJ \ mol^{-1}$.
Step-$2$: $1 \ mol \ H_2O(l, 0^{\circ} C) \rightarrow 1 \ mol \ H_2O(s, 0^{\circ} C)$,Enthalpy change $\Delta H_2 = -6.00 \ kJ \ mol^{-1}$ (freezing is the reverse of fusion).
Total enthalpy change: $\Delta H = \Delta H_1 + \Delta H_2 = -7.56 \ kJ \ mol^{-1} - 6.00 \ kJ \ mol^{-1} = -13.56 \ kJ \ mol^{-1}$.
Since the change in volume during the phase transition from liquid to solid is negligible,$P\Delta V \approx 0$.
Using the relation $\Delta H = \Delta U + P\Delta V$,we get $\Delta U = \Delta H = -13.56 \ kJ \ mol^{-1}$.

(N/A) The formation reaction of benzene is given by:
$6 \, C(graphite) + 3 \, H_{2(g)} \rightarrow C_6H_{6(l)}; \Delta_f H^{\ominus} = ? \dots (i)$
The enthalpy of combustion of $1 \, mol$ of benzene is:
$C_6H_{6(l)} + \frac{15}{2} O_{2(g)}$ $\rightarrow 6 \, CO_{2(g)} + 3 \, H_2O_{(l)}; \Delta_c H^{\ominus} = -3267.0 \, kJ \, mol^{-1} \dots (ii)$
The enthalpy of formation of $1 \, mol$ of $CO_{2(g)}$ is:
$C(graphite) + O_{2(g)} \rightarrow CO_{2(g)}; \Delta_f H^{\ominus} = -393.5 \, kJ \, mol^{-1} \dots (iii)$
The enthalpy of formation of $1 \, mol$ of $H_2O_{(l)}$ is:
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)}; \Delta_f H^{\ominus} = -285.83 \, kJ \, mol^{-1} \dots (iv)$
Using Hess's Law,the enthalpy of combustion is given by:
$\Delta_c H^{\ominus} = [6 \times \Delta_f H^{\ominus}(CO_{2(g)}) + 3 \times \Delta_f H^{\ominus}(H_2O_{(l)})] - [\Delta_f H^{\ominus}(C_6H_{6(l)}) + \frac{15}{2} \Delta_f H^{\ominus}(O_{2(g)})]$
Since $\Delta_f H^{\ominus}(O_{2(g)}) = 0$,we have:
$-3267.0 = [6 \times (-393.5) + 3 \times (-285.83)] - \Delta_f H^{\ominus}(C_6H_{6(l)})$
$-3267.0 = [-2361.0 - 857.49] - \Delta_f H^{\ominus}(C_6H_{6(l)})$
$-3267.0 = -3218.49 - \Delta_f H^{\ominus}(C_6H_{6(l)})$
$\Delta_f H^{\ominus}(C_6H_{6(l)}) = -3218.49 + 3267.0 = 48.51 \, kJ \, mol^{-1}$

The enthalpy change for a reaction $(\Delta H)$ is given by the expression:
$\Delta H = \Delta U + \Delta n_g RT$
Where:
$\Delta U = -742.7 \ kJ \ mol^{-1}$ (change in internal energy)
$\Delta n_g = \Sigma n_g(\text{products}) - \Sigma n_g(\text{reactants})$
For the reaction: $NH_2CN_{(s)} + 1.5 \ O_{2_{(g)}} \to N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$
$\Delta n_g = (1 + 1) - 1.5 = 0.5 \ mol$
$T = 298 \ K$
$R = 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$
Substituting the values:
$\Delta H = -742.7 + (0.5 \times 8.314 \times 10^{-3} \times 298)$
$\Delta H = -742.7 + 1.239$
$\Delta H = -741.46 \ kJ \ mol^{-1}$

The total enthalpy change is the sum of the following three steps:
$(a)$ Cooling $1 \ mol$ of water from $10^{\circ} C$ to $0^{\circ} C$: $\Delta H_1 = n C_p [H_2 O_{(l)}] \Delta T = 1 \ mol \times 75.3 \ J \ mol^{-1} \ K^{-1} \times (0 - 10) \ K = -753 \ J$.
$(b)$ Freezing $1 \ mol$ of water at $0^{\circ} C$ to ice at $0^{\circ} C$: $\Delta H_2 = -\Delta_{fus} H = -6.03 \ kJ \ mol^{-1} = -6030 \ J$.
$(c)$ Cooling $1 \ mol$ of ice from $0^{\circ} C$ to $-10^{\circ} C$: $\Delta H_3 = n C_p [H_2 O_{(s)}] \Delta T = 1 \ mol \times 36.8 \ J \ mol^{-1} \ K^{-1} \times (-10 - 0) \ K = -368 \ J$.
Total $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -753 \ J - 6030 \ J - 368 \ J = -7151 \ J = -7.151 \ kJ \ mol^{-1}$.

(N/A) For the reaction $2A_{(g)} + B_{(g)} \to 2D_{(g)}$,the change in the number of gaseous moles is $\Delta n_{g} = 2 - (2 + 1) = -1 \ mol$.
First,calculate $\Delta H^{\theta}$ using the relation $\Delta H^{\theta} = \Delta U^{\theta} + \Delta n_{g} R T$:
$\Delta H^{\theta} = -10.5 \ kJ + (-1 \ mol)(8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1})(298 \ K)$
$\Delta H^{\theta} = -10.5 \ kJ - 2.477 \ kJ = -12.977 \ kJ$.
Now,calculate $\Delta G^{\theta}$ using $\Delta G^{\theta} = \Delta H^{\theta} - T \Delta S^{\theta}$:
$\Delta G^{\theta} = -12.977 \ kJ - (298 \ K)(-44.1 \times 10^{-3} \ kJ \ K^{-1})$
$\Delta G^{\theta} = -12.977 \ kJ + 13.142 \ kJ = +0.165 \ kJ$.
Since $\Delta G^{\theta} > 0$,the reaction is non-spontaneous at $298 \ K$.

(A) The combustion reaction for liquid benzene $(C_6H_6)$ is: $C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$.
The change in the number of moles of gas is $\Delta n_g = n_p(g) - n_r(g) = 6 - 7.5 = -1.5$.
The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is $\Delta H = \Delta U + \Delta n_g RT$.
Given $\Delta U = -3268 \ kJ \ mol^{-1} = -3268000 \ J \ mol^{-1}$,$T = 300 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$\Delta H = -3268000 + (-1.5 \times 8.314 \times 300)$.
$\Delta H = -3268000 - 3741.3 \ J \ mol^{-1} = -3271741.3 \ J \ mol^{-1}$.
Therefore,$\Delta H = -3271.74 \ kJ \ mol^{-1}$.

(A) The neutralization reaction is: $H^+_{(aq)} + OH^-_{(aq)} \rightleftharpoons H_2O_{(l)}$
The enthalpy change for this reaction is $\Delta H = -13.7 \ k.cal/mol$.
Using the formula: $\Delta H = \Delta_f H^{\ominus}(H_2O) - [\Delta_f H^{\ominus}(H^+) + \Delta_f H^{\ominus}(OH^-)]$
Given $\Delta_f H^{\ominus}(H^+) = 0 \ k.cal/mol$ (by convention) and $\Delta_f H^{\ominus}(H_2O) = -68 \ k.cal/mol$.
Substituting the values: $-13.7 = -68 - [0 + \Delta_f H^{\ominus}(OH^-)]$
$\Delta_f H^{\ominus}(OH^-) = -68 + 13.7 = -54.3 \ k.cal/mol$.

(B) $1$. Calculate $\Delta n_g$: $\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - (2 + 1) = -1 \ mol$.
$2$. Calculate $\Delta H^o$: $\Delta H^o = \Delta U + \Delta n_g RT = -10.5 \ kJ + (-1 \ mol \times 8.314 \times 10^{-3} \ kJ/K \cdot mol \times 298 \ K) = -10.5 - 2.477 = -12.977 \ kJ$.
$3$. Calculate $\Delta G^o$: $\Delta G^o = \Delta H^o - T\Delta S^o = -12.977 \ kJ - (298 \ K \times -10.5 \times 10^{-3} \ kJ/K) = -12.977 + 3.129 = -9.848 \ kJ$.
$4$. Since $\Delta G^o < 0$,the reaction is spontaneous.

(A) The correct matches are:
$(a-5)$ Adiabatic process involves no heat exchange.
$(b-4)$ Isolated system allows no exchange of matter or energy.
$(c-6)$ Isothermal change occurs at constant temperature.
$(d-1)$ Heat is a path function.
$(e-7, 11)$ Internal energy and entropy are state functions.
$(f-2)$ $\Delta U = q$ at constant volume.
$(g-3)$ Law of conservation of energy is the first law of thermodynamics.
$(h-10)$ Reversible process is infinitely slow and involves multiple equilibrium states.
$(i-8)$ Free expansion occurs when $p_{ext} = 0$.
$(j-9)$ $\Delta H = q$ at constant pressure.
$(k-12)$ Pressure is an intensive property.
$(l-7, 11)$ Internal energy and entropy are extensive properties (Note: Specific heat is intensive).

(N/A) $(i)$ Conservation of energy
$(ii)$ Zeroth
$(iii)$ First
$(iv)$ Heat capacity

(A) The correct matches are:
$(a-5)$ (Adiabatic process: No heat exchange)
$(b-4)$ (Isolated system: No exchange of matter and energy)
$(c-6)$ (Isothermal change: Constant temperature)
$(d-1)$ (Path function: Heat)
$(e-7, 11)$ (State function: Internal energy,Entropy)
$(f-2)$ ($\Delta U = q$ at constant volume)
$(g-3)$ (Law of conservation of energy: First law of thermodynamics)
$(h-10)$ (Reversible process: Infinitely slow process)
$(i-8)$ (Free expansion: $p_{ext} = 0$)
$(j-9)$ ($\Delta H = q$ at constant pressure)
$(k-12, 13)$ (Intensive property: Pressure,Specific heat)
$(l-7, 11)$ (Extensive property: Internal energy,Entropy)

(A) The correct matches are:
$(a)$ Liquid to vapor conversion involves an increase in disorder,so $\Delta S = (+)$.
$(b)$ $A$ process that is not spontaneous at any temperature typically involves a decrease in entropy and an increase in enthalpy,so $\Delta S = (-)$.
$(c)$ For a reversible process in an isolated system or specific reversible expansions,the entropy change of the universe is zero,so $\Delta S = 0$.
Therefore,the correct sequence is $(a-2, b-3, c-1)$.

(A) The correct matches are:
$(a)$ Entropy of vaporization is defined as $\frac{\Delta H_{vap}}{T_b}$ (Option $4$).
$(b)$ For a spontaneous process,the equilibrium constant $K > 1$,which implies $\Delta G < 0$,and it is associated with a positive value of entropy change in the universe (Option $2$).
$(c)$ Crystalline solid state has the most ordered arrangement,thus it has minimum entropy (Option $3$).
$(d)$ For adiabatic expansion of an ideal gas,$\Delta U = q + w$. Since $q = 0$,$\Delta U = w$. As the gas expands,work is done by the system,so $\Delta U$ decreases (Option $1$).
Therefore,the correct sequence is $a-4, b-2, c-3, d-1$.

(A) The reaction for evaporation is: $H_2O(\ell) \rightleftharpoons H_2O(g)$.
Moles of water $(n)$ = $\frac{90 \ g}{18 \ g/mol} = 5 \ mol$.
The relationship between enthalpy change and internal energy change is: $\Delta H = \Delta U + \Delta n_g RT$.
Total enthalpy change $(\Delta H)$ = $n \times \Delta H_{vap} = 5 \ mol \times 41000 \ J/mol = 205000 \ J$.
Change in moles of gas $(\Delta n_g)$ = $5 \ mol$ (since $5 \ mol$ of liquid water produces $5 \ mol$ of water vapor).
Substituting the values: $205000 \ J = \Delta U + (5 \ mol \times 8.314 \ J \ K^{-1} mol^{-1} \times 373 \ K)$.
$205000 = \Delta U + 15505.59$.
$\Delta U = 205000 - 15505.59 = 189494.41 \ J$.

(B) The reaction is $2 A_{(g)} \rightarrow A_{2(g)}$.
The change in the number of gaseous moles is $\Delta n_g = 1 - 2 = -1$.
First,calculate $\Delta H^{\ominus}$ using the relation $\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT$.
$\Delta H^{\ominus} = -20 \ kJ \ mol^{-1} + (-1) \times (8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) \times 298 \ K$.
$\Delta H^{\ominus} = -20 - 2.477572 = -22.477572 \ kJ \ mol^{-1}$.
Now,calculate $\Delta G^{\ominus}$ using $\Delta G^{\ominus} = \Delta H^{\ominus} - T \Delta S^{\ominus}$.
$\Delta G^{\ominus} = -22.477572 \ kJ \ mol^{-1} - (298 \ K \times -30 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1})$.
$\Delta G^{\ominus} = -22.477572 + 8.94 = -13.537572 \ kJ \ mol^{-1}$.
Converting to Joules: $\Delta G^{\ominus} = -13.537572 \times 1000 \ J \ mol^{-1} = -13537.57 \ J \ mol^{-1}$.

(A) In the reaction $2 Cl ( g ) \rightarrow Cl _{2}( g )$,two gaseous atoms combine to form one molecule of chlorine gas.
$1$. Enthalpy change $(\Delta_{ r } H)$: Bond formation is an exothermic process,therefore energy is released,making $\Delta_{ r } H < 0$.
$2$. Entropy change $(\Delta_{ r } S)$: The number of moles of gas decreases from $2$ to $1$ $(2 Cl ( g ) \rightarrow 1 Cl _{2}( g ))$. Since the randomness or disorder of the system decreases,$\Delta_{ r } S < 0$.
Thus,both $\Delta_{ r } H < 0$ and $\Delta_{ r } S < 0$.

(D) For any chemical reaction,the relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by:
$\Delta H = \Delta E + \Delta n_{g}RT$
Where $\Delta n_{g}$ is the change in the number of moles of gaseous species.
$\Delta n_{g} = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
For the reaction: $Fe_{2}O_{3(s)} + 3H_{2(g)} \rightarrow 2Fe_{(s)} + 3H_{2}O_{(l)}$
Note that $Fe_{2}O_{3}$ and $Fe$ are solids,and $H_{2}O$ is a liquid. Only $H_{2}$ is in the gaseous state.
$\Delta n_{g} = 0 - 3 = -3$
Substituting this value into the equation:
$\Delta H = \Delta E + (-3)RT$
$\Delta H = \Delta E - 3RT$

(B) The combustion reaction of butane is: $C_4H_{10(g)} + \frac{13}{2} O_{2(g)} \rightarrow 4 CO_{2(g)} + 5 H_2O_{(l)}$
The change in the number of moles of gaseous species is calculated as:
$\Delta n_g = n_{g(products)} - n_{g(reactants)} = 4 - (1 + 6.5) = 4 - 7.5 = -3.5$
The work done $(W)$ for a chemical reaction is given by the formula:
$W = -\Delta n_g RT$
Given:
$R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$
$T = 25 + 273 = 298 \text{ K}$
Substituting the values:
$W = -(-3.5 \times 0.0821 \times 298)$
$W = 3.5 \times 0.0821 \times 298$
$W = 85.63 \text{ L atm}$
Thus,the work done is approximately $85.6 \text{ L atm}$.

(A) The chemical reaction is: $Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)$.
Moles of $Fe = \frac{50 \ g}{55.85 \ g \ mol^{-1}} \approx 0.89526 \ mol$.
Since $1 \ mol$ of $Fe$ produces $1 \ mol$ of $H_2(g)$,moles of $H_2(g)$ produced $(n)$ $= 0.89526 \ mol$.
Work done $(w)$ $= -P_{ext} \Delta V = -\Delta n_g RT$.
Given $T = 25^{\circ} C = 298 \ K$,$R = 8.314 \ J \ mol^{-1} \ K^{-1}$,and $P_{ext} = 1 \ bar$.
$w = -nRT = -0.89526 \ mol \times 8.314 \ J \ mol^{-1} \ K^{-1} \times 298 \ K$.
$w = -2218.059 \ J$.
The magnitude of work done by the gas is $2218 \ J$.

(B) The reaction is: $NH_{2}CN_{(s)} + \frac{3}{2} O_{2(g)} \rightarrow N_{2(g)} + CO_{2(g)} + H_{2}O_{(l)}$
The change in the number of moles of gaseous species is $\Delta n_{g} = (n_{products, g} - n_{reactants, g}) = (1 + 1) - \frac{3}{2} = 2 - 1.5 = 0.5 \ mol$.
Using the relation $\Delta H = \Delta U + \Delta n_{g} RT$:
$\Delta H = -742.24 \ kJ \ mol^{-1} + (0.5 \ mol) \times (8.314 \times 10^{-3} \ kJ \ mol^{-1} K^{-1}) \times (298 \ K)$
$\Delta H = -742.24 + 1.238786 \approx -741.001 \ kJ \ mol^{-1}$.
The magnitude of $\Delta H$ is $741 \ kJ \ mol^{-1}$.

(C) $\text{Millimoles of } HCl = 200 \times 0.2 = 40 \ mmol$
$\text{Millimoles of } NaOH = 300 \times 0.1 = 30 \ mmol$
$\text{Since } NaOH \text{ is the limiting reagent, heat released } (q) = \frac{30}{1000} \times 57.1 \times 1000 = 1713 \ J$
$\text{Total mass of solution } (m) = (200 + 300) \ mL \times 1 \ g \ mL^{-1} = 500 \ g$
$\Delta T = \frac{q}{m \times C} = \frac{1713}{500 \times 4.18} = \frac{1713}{2090} \approx 0.8196 \ K$
$\Delta T = 81.96 \times 10^{-2} \ K$
$\text{Since } \Delta T \text{ in } K \text{ is equal to } \Delta T \text{ in } ^{\circ}C, \text{ we have } x = 81.96 \approx 82$.

(C) For an ideal gas,the internal energy $ \Delta U $ depends only on temperature. Since the process is isothermal $( \Delta T = 0 )$,$ \Delta U = 0 $.
For any irreversible process,the total entropy change of the universe is always positive,i.e.,$ \Delta S_{total} > 0 $.
Therefore,$ \Delta S_{total} \neq 0 $.
Thus,the correct condition is $ \Delta U = 0 $ and $ \Delta S_{total} \neq 0 $.

(A) In a $p-V$ graph,the area under the curve represents the magnitude of the work done.
Comparing the given graphs,the area enclosed under the curve in option $A$ is the largest among the choices provided.
Therefore,the curve in option $A$ represents the maximum work done.

(C) The enthalpy of reaction is calculated using the enthalpies of combustion of reactants and products:
$\Delta H = \sum \Delta H_{\text{combustion}} \text{ (Reactants)} - \sum \Delta H_{\text{combustion}} \text{ (Products)}$
For the reaction $3 C_2H_{2(g)} \rightarrow C_6H_{6(l)}$:
$\Delta H = [3 \times \Delta H_{\text{combustion}}(C_2H_{2(g)})] - [1 \times \Delta H_{\text{combustion}}(C_6H_{6(l)})]$
Substituting the given values:
$\Delta H = [3 \times (-1300 \, kJ \, mol^{-1})] - [-3268 \, kJ \, mol^{-1}]$
$\Delta H = -3900 + 3268 = -632 \, kJ \, mol^{-1}$

(A) The vaporization process is represented as: $H_2O_{(l)} \rightarrow H_2O_{(g)}$.
The number of moles of water is $n = \frac{36 \ g}{18 \ g \ mol^{-1}} = 2 \ mol$.
The relationship between enthalpy change and internal energy change is given by: $\Delta U = \Delta H - \Delta n_g RT$.
For the vaporization of $1 \ mol$ of water,$\Delta n_g = 1$ (since $1 \ mol$ of gas is produced from $1 \ mol$ of liquid).
$\Delta U = \Delta_{vap} H^{\ominus} - RT = 41.1 \ kJ \ mol^{-1} - (8.31 \ J \ K^{-1} \ mol^{-1} \times 373 \ K) / 1000$.
$\Delta U = 41.1 - 3.09963 \approx 41.1 - 3.1 = 38.0 \ kJ \ mol^{-1}$.
Thus,the internal energy change for vaporization is $38 \ kJ \ mol^{-1}$.

(D) The heat measured by a bomb calorimeter is the change in internal energy,$\Delta U = -726 \ kJ \ mol^{-1}$.
The reaction is $CH_3OH_{(l)} + \frac{3}{2} O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$.
The change in the number of gaseous moles is $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - \frac{3}{2} = -0.5 \ mol$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Given $T = 27 + 273 = 300 \ K$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1} = 8.3 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
$\Delta H = -726 + (-0.5) \times (8.3 \times 10^{-3}) \times 300$.
$\Delta H = -726 - 1.245 = -727.245 \ kJ \ mol^{-1}$.
Rounding to the nearest integer,$x = 727$.

(B) . For a spontaneous process,$\Delta G_{T, P} < 0$. Thus,$A-II$.
$B$. $\Delta P = 0$ represents an isobaric process and $\Delta T = 0$ represents an isothermal process. Thus,$B-III$.
$C$. $\Delta H_{reaction} = (\Sigma \text{Bond energies of reactants}) - (\Sigma \text{Bond energies of products})$. Thus,$C-IV$.
$D$. An exothermic process is characterized by $\Delta H < 0$. Thus,$D-I$.
Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.

(D) $1$. Calculate the number of moles of $N_{2}O$: $n = \frac{2.2 \, g}{44 \, g \, mol^{-1}} = 0.05 \, mol$.
$2$. Calculate the change in enthalpy $(\Delta H)$ at constant pressure: $\Delta H = n C_{p} \Delta T = 0.05 \, mol \times 100 \, J \, K^{-1} \, mol^{-1} \times (270 - 310) \, K = 5 \times (-40) = -200 \, J$.
$3$. Calculate the work done $(w)$ during the process: $w = -P_{ext} \Delta V = -1 \, atm \times (167.75 - 217.1) \, mL = -1 \times (-49.35) \, mL \cdot atm = 49.35 \, mL \cdot atm$.
$4$. Convert $w$ to Joules: $1 \, L \cdot atm = 101.3 \, J$,so $w = 49.35 \times 10^{-3} \, L \times 101.3 \, J \, L^{-1} \approx 5 \, J$.
$5$. Use the relation $\Delta H = \Delta U + P \Delta V$,which implies $\Delta U = \Delta H - P \Delta V = \Delta H + w$.
$6$. $\Delta U = -200 \, J + 5 \, J = -195 \, J$.
$7$. Since $\Delta U = -x \, J$,we have $-195 = -x$,so $x = 195$.

(D) The combustion reaction is: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$; $\Delta H = -x \ kJ \ mol^{-1}$.
Heat released $(Q)$ is calculated as: $Q = C \times \Delta T = 20.0 \ kJ \ K^{-1} \times (300 \ K - 298 \ K) = 20.0 \ kJ \ K^{-1} \times 2 \ K = 40 \ kJ$.
This heat is released by $2.4 \ g$ of carbon $(C)$.
For $1 \ mole$ of $C$ $(12 \ g)$:
$Q = \frac{40 \ kJ}{2.4 \ g} \times 12 \ g = 200 \ kJ \ mol^{-1}$.
Since $\Delta n_g = 0$ for the reaction,$\Delta H = \Delta U = -200 \ kJ \ mol^{-1}$.
Therefore,$x = 200$.

(A) Let $x \ g$ be the mass of the gas burnt.
Moles of gas $= \frac{x}{280} \ mol$.
Heat released by the combustion $= \text{Heat capacity} \times \Delta T = 2.5 \ kJ \ K^{-1} \times (298.45 - 298.0) \ K = 2.5 \times 0.45 \ kJ = 1.125 \ kJ$.
Since the combustion occurs in a constant volume calorimeter,the heat released is equal to the internal energy change,$\Delta U$.
Given the enthalpy of combustion $\Delta H = 9 \ kJ \ mol^{-1}$,and assuming $\Delta H \approx \Delta U$ for this calculation,we have:
$\Delta U \text{ per mole} = 9 \ kJ \ mol^{-1}$.
Total heat released $= \text{moles} \times \Delta U \text{ per mole}$.
$1.125 \ kJ = (\frac{x}{280} \ mol) \times 9 \ kJ \ mol^{-1}$.
$x = \frac{1.125 \times 280}{9} = 35 \ g$.

(B) For an ideal gas,the relationship between molar heat capacities is $C_{p,m} - C_{v,m} = R$.
Given $C_{p,m} = 20.785 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,we calculate $C_{v,m}$ as:
$C_{v,m} = 20.785 - 8.314 = 12.471 \ J \ K^{-1} \ mol^{-1}$.
The change in internal energy $\Delta U$ is given by the formula $\Delta U = n C_{v,m} \Delta T$.
Here,$\Delta U = 5000 \ J$ and $\Delta T = 500 \ K - 300 \ K = 200 \ K$.
Substituting the values: $5000 = n \times 12.471 \times 200$.
$n = \frac{5000}{12.471 \times 200} = \frac{25}{12.471} \approx 2.0046$.
The nearest integer is $2$.

(A) By definition,the enthalpy change is given by $\Delta H = \Delta U + \Delta(PV)$.
At constant pressure,this becomes $\Delta H = \Delta U + P \Delta V$.
Option $A$ is written as $\Delta H = \Delta U - P \Delta V$,which is incorrect because the correct sign is positive.
Option $B$ represents the first law of thermodynamics $(\Delta U = q + W)$.
Option $C$ represents the second law of thermodynamics for a spontaneous process.
Option $D$ is the Gibbs-Helmholtz equation.

(D) Given equations:
$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} + 400 \; kJ \; (I)$
$C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} + 100 \; kJ \; (II)$
Mass of coal $= 0.6 \; kg = 600 \; g$
Pure Carbon $= 600 \times \frac{60}{100} = 360 \; g$
Moles of pure Carbon $= \frac{360 \; g}{12 \; g/mol} = 30 \; mol$
Carbon converted into $CO = 30 \times 0.60 = 18 \; mol$
Carbon converted into $CO_2 = 30 - 18 = 12 \; mol$
Energy from $CO$ formation $= 18 \; mol \times 100 \; kJ/mol = 1800 \; kJ$
Energy from $CO_2$ formation $= 12 \; mol \times 400 \; kJ/mol = 4800 \; kJ$
Total heat generated $= 1800 \; kJ + 4800 \; kJ = 6600 \; kJ$

(B) For a chemical reaction to be spontaneous at constant temperature and pressure,the change in Gibbs energy of the system must be negative,i.e.,$\Delta G_{\text{sys}} < 0$. This corresponds to statement $(i)$.
According to the second law of thermodynamics,for any spontaneous process,the total entropy change of the universe (system + surroundings) must be positive,i.e.,$\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$. This corresponds to statement $(iv)$.
Statement $(ii)$ is incorrect because the total Gibbs energy change for the universe is not a standard criterion for spontaneity.
Statement $(iii)$ is not necessarily true because a reaction can be spontaneous even if the entropy change of the system is negative,provided the enthalpy change is sufficiently negative to make $\Delta G$ negative.
Therefore,statements $(i)$ and $(iv)$ are always true.

(D) For a cyclic process,the change in internal energy $\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = q + W = 0$,which implies $q = -W$.
Total work done $W_{total} = W_{AB} + W_{BC} + W_{CA}$.
Process $A \rightarrow B$: Isochoric (constant volume),so $W_{AB} = 0$.
Process $B \rightarrow C$: Given $W_{BC} = 1 \, L \, atm$.
Process $C \rightarrow A$: Isobaric (constant pressure at $2 \, atm$),so $W_{CA} = -p \Delta V = -2 \, atm \times (1 \, L - 1.5 \, L) = -2 \times (-0.5) = 1 \, L \, atm$.
Total work $W_{total} = 0 + 1 + 1 = 2 \, L \, atm$.
Since $q = -W_{total}$,the heat exchanged is $q = -2 \, L \, atm$.
Re-evaluating the provided options and the cyclic nature,the net heat exchanged in a cycle is equal to the negative of the total work done. Based on the graph,the area enclosed is the work. The area of the triangle $ABC$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2-1) \, atm \times (1.5-1) \, L = 0.25 \, L \, atm$.
Given the specific path $A$ $\rightarrow B$ $\rightarrow C$ $\rightarrow A$,the net work is $W_{AB} + W_{BC} + W_{CA} = 0 + 1 + 1 = 2 \, L \, atm$.
There appears to be a discrepancy in the provided options. Based on standard thermodynamic cycles,the correct magnitude is $2.0 \, L \, atm$.

(D) The vaporization reaction is: $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$.
Given: $\Delta H = 41 \ kJ \ mol^{-1}$,$T = 373 \ K$,$R = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Here,$\Delta n_g = n_g(\text{products}) - n_g(\text{reactants}) = 1 - 0 = 1$.
Substituting the values: $41 = \Delta U + (1 \times 8.314 \times 10^{-3} \times 373)$.
$41 = \Delta U + 3.101$.
$\Delta U = 41 - 3.101 = 37.899 \approx 37.9 \ kJ \ mol^{-1}$.

(B)
Given,$\Delta H_{vap} = 20 \, kJ/mol$.
For the process of vaporisation,the change in the number of moles of gas is $\Delta n_g = n_{gas} - n_{liquid} \approx 1 - 0 = 1$.
Temperature $T = 60 + 273 = 333 \, K$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging for internal energy change: $\Delta U = \Delta H - \Delta n_g RT$.
Substituting the values: $\Delta U = 20 \, kJ/mol - (1 \, mol \times 8.314 \times 10^{-3} \, kJ \cdot K^{-1} \cdot mol^{-1} \times 333 \, K)$.
$\Delta U = 20 - 2.768 \approx 17.23 \, kJ/mol$.
Thus,the value is close to $17.2 \, kJ/mol$.

(A) The thermodynamic cycle in the $p-V$ plane consists of two isothermal and two adiabatic processes.
$1. A \rightarrow B$: Isothermal expansion $(T = \text{constant})$,entropy increases ($S$ increases).
$2. B \rightarrow C$: Adiabatic expansion $(S = \text{constant})$,temperature decreases ($T$ decreases).
$3. C \rightarrow D$: Isothermal compression $(T = \text{constant})$,entropy decreases ($S$ decreases).
$4. D \rightarrow A$: Adiabatic compression $(S = \text{constant})$,temperature increases ($T$ increases).
In the $T-S$ plane,isothermal processes appear as horizontal lines $(T = \text{constant})$ and adiabatic processes appear as vertical lines $(S = \text{constant})$. Based on the directions of the processes,the correct representation is a rectangle in the $T-S$ plane with $A \rightarrow B$ at a higher temperature than $C \rightarrow D$.

(D) The total heat required $(q)$ is the sum of the heat required to raise the temperature of the liquid $(q_1)$ and the heat required for phase change $(q_2)$.
$q_1 = m \times c \times \Delta T$
$q_1 = 1000 \, g \times 2.44 \, J \, g^{-1} \, K^{-1} \times (351.45 \, K - 293.45 \, K)$
$q_1 = 1000 \times 2.44 \times 58 = 141520 \, J = 1.4152 \times 10^5 \, J$
$q_2 = m \times L_v$
$q_2 = 1000 \, g \times 855 \, J \, g^{-1} = 855000 \, J = 8.55 \times 10^5 \, J$
$q = q_1 + q_2 = 1.4152 \times 10^5 \, J + 8.55 \times 10^5 \, J = 9.9652 \times 10^5 \, J$
Rounding to the nearest value,we get $9.97 \times 10^5 \, J$.

(D) The heat gained by the ice equals the heat lost by the water.
Heat gained by $X \, g$ of ice at $0^{\circ} C$ to become water at $5^{\circ} C$ is given by: $Q_{gain} = X \times L_f + X \times c \times \Delta T_{ice}$.
$Q_{gain} = X(333) + X(4.184)(5 - 0) = 333X + 20.92X = 353.92X$.
Heat lost by $340 \, g$ of water cooling from $20^{\circ} C$ to $5^{\circ} C$ is given by: $Q_{lost} = m \times c \times \Delta T_{water}$.
$Q_{lost} = 340 \times 4.184 \times (20 - 5) = 340 \times 4.184 \times 15 = 21338.4 \, J$.
Equating $Q_{gain} = Q_{lost}$:
$353.92X = 21338.4$.
$X = \frac{21338.4}{353.92} \approx 60.29 \, g$.
Thus,the value of $X$ is approximately $60.3 \, g$.

(A) For an ideal gas in an isolated system,the process of mixing is spontaneous and occurs at constant temperature (isothermal process).
Since the system is isolated,there is no exchange of heat with the surroundings $(q = 0)$ and no work is done $(w = 0)$,so $\Delta U = 0$.
For an ideal gas,enthalpy change is given by $\Delta H = \Delta U + \Delta(PV) = \Delta U + \Delta(nRT)$. Since $T$ is constant and $\Delta U = 0$,$\Delta H = 0$.
When the partition is removed,the gases mix,which increases the randomness of the system. Therefore,the entropy of the system increases,meaning $\Delta S > 0$ (positive).
Thus,the change in enthalpy is zero and the change in entropy is positive.