Free energy and Work Questions in English

(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative. The relationship is given by $\Delta G = \Delta H - T \cdot \Delta S$.
If $\Delta H$ is negative (exothermic) and $\Delta S$ is positive (increase in entropy),then $\Delta G$ will be negative at all temperatures $T$ because both terms $\Delta H$ and $-T \cdot \Delta S$ will be negative.

(B) The Gibbs free energy is defined as $G = H - TS$.
For a system at constant pressure,the change in Gibbs free energy with respect to temperature is given by the fundamental thermodynamic relation:
$[dG/dT]_P = -S$.
Thus,the correct relation is $[dG/dT]_P = -S$.

(A) From the fundamental thermodynamic relation for Gibbs free energy,$dG = VdP - SdT$.
At constant pressure,$dP = 0$,so $dG = -SdT$.
Therefore,${\left[ {\frac{dG}{dT}} \right]_P} = -S$.

(C) The Gibbs free energy equation is given by: $\Delta G = \Delta H - T \cdot \Delta S$.
Given that $\Delta H = \Delta S$,we substitute $\Delta S$ with $\Delta H$ in the equation:
$\Delta G = \Delta H - T \cdot \Delta H$.
Factoring out $\Delta H$:
$\Delta G = \Delta H(1 - T)$.
Rearranging to solve for $\Delta H$:
$\Delta H = \frac{\Delta G}{1 - T}$.

(A) The Gibbs free energy equation is given by $\Delta G = \Delta H - T \cdot \Delta S$.
For a reaction to be non-spontaneous,the change in Gibbs free energy must be positive,i.e.,$\Delta G > 0$.
Substituting the expression,we get $\Delta H - T \cdot \Delta S > 0$,which implies $\Delta H > T \cdot \Delta S$.

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G < 0)$.
Given the equation $\Delta G = \Delta H - T\Delta S$,for $\Delta G < 0$,we have $\Delta H - T\Delta S < 0$.
This implies $\Delta H < T\Delta S$,or $T > \frac{\Delta H}{\Delta S}$.
Given $\Delta H = 3 \, kJ = 3000 \, J$ and $\Delta S = 10 \, J/K$.
Substituting these values: $T > \frac{3000 \, J}{10 \, J/K} = 300 \, K$.
Thus,the reaction becomes spontaneous at temperatures above $300 \, K$.

(A) The Gibbs free energy equation is given by $\Delta G = \Delta H - T \cdot \Delta S$.
For a reaction to be spontaneous,$\Delta G$ must be less than $0$.
Therefore,$\Delta H - T \cdot \Delta S < 0$,which implies $\Delta H < T \cdot \Delta S$.
Since both $\Delta H$ and $\Delta S$ are positive,the condition $\Delta H < T \cdot \Delta S$ is satisfied only when the temperature $T$ is high.
Thus,the reaction is spontaneous at high temperatures.

(C) For a spontaneous reaction,the Gibbs free energy change $\Delta G$ must be negative,i.e.,$\Delta G < 0$.
Given $\Delta G = \Delta H - T\Delta S < 0$,which implies $\Delta H < T\Delta S$ or $T > \frac{\Delta H}{\Delta S}$.
Convert $\Delta H$ to calories: $\Delta H = 4 \, kcal \, mol^{-1} = 4000 \, cal \, mol^{-1}$.
Calculate the equilibrium temperature: $T = \frac{\Delta H}{\Delta S} = \frac{4000 \, cal \, mol^{-1}}{10 \, cal \, K^{-1} \, mol^{-1}} = 400 \, K$.
Since $T > 400 \, K$,the reaction is spontaneous at temperatures greater than $400 \, K$. Among the given options,$500 \, K$ is the correct temperature where the reaction is spontaneous.

(D) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For a reaction to be spontaneous at all temperatures,$\Delta G$ must be negative.
This occurs when the enthalpy change $\Delta H$ is negative (exothermic) and the entropy change $\Delta S$ is positive (increase in disorder).
In this case,$\Delta G = (\text{negative}) - T(\text{positive})$,which will always be negative regardless of the temperature $T$.

(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,where $\Delta G = \Delta H - T\Delta S$.
For endothermic reactions,$\Delta H > 0$. If the reaction involves an increase in entropy $(\Delta S > 0)$,then at high temperatures,the term $T\Delta S$ becomes larger than $\Delta H$.
Consequently,$\Delta G = \Delta H - T\Delta S$ becomes negative,making the reaction spontaneous at high temperatures.
The reason provided is incorrect because the entropy of a system is generally considered a state function that does not necessarily increase with temperature in the context of the spontaneity condition; rather,the term $T\Delta S$ increases with temperature.

(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative.
The relationship is given by $\Delta G = \Delta H - T\Delta S$.
According to the reaction $A + B \rightarrow C + D + q$,the term $q$ indicates that heat is evolved,so the enthalpy change $\Delta H$ is negative.
Given that the entropy change $\Delta S$ is positive,the term $-T\Delta S$ will always be negative for any temperature $T > 0$.
Since both $\Delta H$ and $-T\Delta S$ are negative,$\Delta G$ will be negative at all temperatures.
Therefore,the reaction is spontaneous at any temperature.

(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative.
The relationship is given by $\Delta G = \Delta H - T \Delta S$.
At equilibrium,$\Delta G = 0$,so $T = \frac{\Delta H}{\Delta S}$.
Substituting the given values: $T = \frac{400 \, kJ \, mol^{-1}}{0.2 \, kJ \, K^{-1} \, mol^{-1}} = 2000 \, K$.
Since $\Delta H$ is positive (endothermic) and $\Delta S$ is positive,the reaction becomes spontaneous when $T \Delta S > \Delta H$.
Therefore,the reaction will be spontaneous at $T > 2000 \, K$.

(N/A) Gibbs energy $(G)$ is an extensive property defined as $G = H - TS$.
The change in Gibbs energy for a system at constant temperature $(\Delta T = 0)$ is given by the Gibbs equation:
$\Delta G = \Delta H - T \Delta S$
Here,$\Delta H$ and $T \Delta S$ are energy terms,so $\Delta G$ has units of energy.
For a system in thermal equilibrium with its surroundings,the temperature of the system equals the temperature of the surroundings. The heat lost by the system is gained by the surroundings,so $\Delta H_{\text{surr}} = -\Delta H_{\text{sys}}$.
The total entropy change is $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = \Delta S_{\text{sys}} - \frac{\Delta H_{\text{sys}}}{T}$.
Multiplying by $-T$,we get $-T \Delta S_{\text{total}} = \Delta H_{\text{sys}} - T \Delta S_{\text{sys}} = \Delta G_{\text{sys}}$.
Criteria for spontaneity:
$1$. If $\Delta G < 0$,the process is spontaneous.
$2$. If $\Delta G > 0$,the process is non-spontaneous.
$3$. If $\Delta G = 0$,the system is at equilibrium.

(N/A) No,it is not possible to decide the spontaneity of a reaction solely from the enthalpy diagram.
Enthalpy change $(\Delta_rH)$ is only one of the contributing factors in determining the spontaneity of a reaction.
Another essential factor,the entropy factor $(\Delta S)$,must also be taken into consideration.
The overall spontaneity is determined by the Gibbs free energy change,given by the equation $\Delta G = \Delta H - T\Delta S$,where a reaction is spontaneous only if $\Delta G < 0$.

(N/A) Gibbs free energy is the thermodynamic quantity of a system,the decrease in whose value during a process is equal to the maximum possible useful work that can be obtained from the system.
Mathematically,this result may be derived as follows:
The relationship between heat absorbed by a system $q$,the change in its internal energy $\Delta U$,and the work done by the system is given by the first law of thermodynamics:
$q = \Delta U + W_{\text{expansion}} + W_{\text{non-expansion}}$ $(i)$
Under constant pressure,the expansion work is $p \Delta V$. Since $\Delta H = \Delta U + p \Delta V$,we have:
$q = \Delta H + W_{\text{non-expansion}}$ $(ii)$
For a reversible process at constant temperature $T$,$\Delta S = \frac{q_{\text{rev}}}{T}$,so $q_{\text{rev}} = T \Delta S$ $(iii)$
Substituting $(iii)$ into $(ii)$:
$T \Delta S = \Delta H + W_{\text{non-expansion}}$
$\Delta H - T \Delta S = -W_{\text{non-expansion}}$ $(iv)$
Since $\Delta G = \Delta H - T \Delta S$,equation $(iv)$ becomes:
$\Delta G = -W_{\text{non-expansion}}$ $(v)$
Thus,$-\Delta G$ represents the maximum useful work (non-expansion work) obtainable from the system.
The unit of $\Delta G$ is the same as energy,which is joule $(J)$ or kilojoule $(kJ)$.
For a reaction with positive $\Delta H$ and positive $\Delta S$,the condition for spontaneity $(\Delta G < 0)$ is:
$\Delta H - T \Delta S < 0$
$T \Delta S > \Delta H$
$T > \frac{\Delta H}{\Delta S}$
Thus,the reaction is spontaneous at high temperatures.

(A) The relationship between Gibbs free energy change $\Delta G$ and the equilibrium constant $K$ is given by $\Delta G = -RT \ln K$.
Given the condition $\Delta H < T \Delta S$,we know that $\Delta G = \Delta H - T \Delta S < 0$.
Since $\Delta G < 0$,the reaction is spontaneous in the forward direction.
Substituting $\Delta G < 0$ into the equation $-RT \ln K < 0$,we get $\ln K > 0$,which implies $K > 1$.

(A) For a process to be spontaneous at constant pressure and temperature,the change in Gibbs free energy must be negative.
$1$. If $\Delta G < 0$,the process is spontaneous.
$2$. If $\Delta G > 0$,the process is non-spontaneous.
$3$. If $\Delta G = 0$,the process is at equilibrium.

(B) The relationship between Gibbs free energy change $(\Delta G)$,enthalpy change $(\Delta H)$,and entropy change $(\Delta S)$ at a constant temperature $(T)$ is given by the Gibbs-Helmholtz equation:
$\Delta G = \Delta H - T \Delta S$

(D) The freezing point of water is $273 \ K$ $(0^{\circ}C)$.
At $100 \ K$,water exists as ice (solid state).
The process of water remaining as ice at $100 \ K$ is spontaneous.
For a spontaneous process,the change in Gibbs free energy,$\Delta G$,is negative $(\Delta G < 0)$.

(A) For a process to be spontaneous at constant pressure and temperature,the change in Gibbs free energy must be negative.
$1$. If $\Delta G < 0$,the process is spontaneous.
$2$. If $\Delta G > 0$,the process is non-spontaneous.
$3$. If $\Delta G = 0$,the process is at equilibrium.

(N/A) The relationship between Gibbs free energy change $(\Delta G)$,enthalpy change $(\Delta H)$,and entropy change $(\Delta S)$ at a constant temperature $(T)$ is given by the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T \Delta S$.

(C) The freezing point of water is $273 \ K$ at $1 \ \text{atm}$ pressure.
At $100 \ K$,the system is below the freezing point of water.
Since the temperature is below the phase transition temperature $(273 \ K)$,the process of freezing (liquid to solid) is spontaneous.
For a spontaneous process,the change in Gibbs free energy is negative,i.e.,$\Delta G < 0$.

(A) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
$1$. If $\Delta H > 0$ (endothermic) and $\Delta S < 0$ (entropy decreases),then $\Delta G$ will always be positive at all temperatures. Thus,the reaction is non-spontaneous at all temperatures. $(a-i)$.
$2$. If $\Delta H < 0$ (exothermic) and $\Delta S < 0$ (entropy decreases),then $\Delta G$ will be negative only at low temperatures (where $|\Delta H| > |T\Delta S|$). Thus,the reaction is spontaneous at low temperatures. $(b-ii)$.
$3$. If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (entropy increases),then $\Delta G$ will always be negative at all temperatures. Thus,the reaction is spontaneous at all temperatures. $(c-iii)$.
Therefore,the correct matching is $(a-i, b-ii, c-iii)$.

(A) The reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ involves the dissociation of one mole of gas into two moles of gas,which leads to an increase in entropy,so $\Delta S > 0$.
Since bond breaking is an endothermic process,the enthalpy change is positive,so $\Delta H > 0$.
For a reaction to be spontaneous,the Gibbs free energy change $\Delta G = \Delta H - T\Delta S$ must be negative.
At high temperatures,the term $T\Delta S$ becomes large and positive,making $\Delta G$ negative when $\Delta H > 0$ and $\Delta S > 0$.

(B) The stability of an oxide is determined by the Gibbs free energy change $(\Delta G)$ of the formation reaction.
For the reaction $4 M (s) + n O_2 (g) \rightarrow 2 M_2 O_n (s)$,the oxide is stable when $\Delta G < 0$.
As the temperature increases,the value of $\Delta G$ increases.
The temperature at which $\Delta G = 0$ is the equilibrium temperature.
Below this temperature,$\Delta G$ is negative,meaning the oxide is stable.
Above this temperature,$\Delta G$ becomes positive,meaning the oxide is unstable and tends to decompose.
Therefore,the point where the free energy change crosses from negative to positive indicates the temperature limit for stability.

(B) For a reaction to be spontaneous,the change in Gibbs free energy,$\Delta_{r} G$,must be negative $(\Delta_{r} G < 0)$.
The relationship is given by the equation: $\Delta_{r} G = \Delta_{r} H - T \Delta_{r} S$.
Substituting the condition for spontaneity: $\Delta_{r} H - T \Delta_{r} S < 0$.
Rearranging the inequality: $\Delta_{r} H < T \Delta_{r} S$ or $\Delta_{r} S > \frac{\Delta_{r} H}{T}$.
Given $\Delta_{r} H = 30 \; kJ \; mol^{-1} = 30,000 \; J \; mol^{-1}$ and $T = 450 \; K$.
$\Delta_{r} S > \frac{30,000 \; J \; mol^{-1}}{450 \; K} = 66.67 \; J \; K^{-1} \; mol^{-1}$.
Therefore,the reaction will be spontaneous if $\Delta_{r} S > 66.67 \; J \; K^{-1} \; mol^{-1}$.

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G^{\circ}$ must be less than $0$.
Given the equation: $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$.
At equilibrium,$\Delta G^{\circ} = 0$,so $T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$.
Given values: $\Delta H^{\circ} = 179.1 \ kJ \ mol^{-1} = 179100 \ J \ mol^{-1}$ and $\Delta S^{\circ} = 160.2 \ J \ K^{-1} \ mol^{-1}$.
Substituting these values into the equation:
$T = \frac{179100 \ J \ mol^{-1}}{160.2 \ J \ K^{-1} \ mol^{-1}} \approx 1117.97 \ K$.
Since the reaction is endothermic and entropy increases,the reaction becomes spontaneous at temperatures above $1118 \ K$.

(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G^{0}$ must be less than $0$.
$\Delta G^{0} = \Delta_{r} H^{0} - T \Delta_{r} S^{0} < 0$
Given $\Delta_{r} H^{0} = 80 \, kJ \, mol^{-1} = 80000 \, J \, mol^{-1}$ and $\Delta_{r} S^{0} = 2T \, J \, K^{-1} \, mol^{-1}$.
Substituting the values:
$80000 - T(2T) < 0$
$80000 - 2T^{2} < 0$
$2T^{2} > 80000$
$T^{2} > 40000$
$T > \sqrt{40000}$
$T > 200 \, K$
Therefore,the minimum temperature at which the reaction becomes spontaneous is $200 \, K$.

(A) For a reaction to be spontaneous,$\Delta G < 0$. Since $\Delta G = \Delta H - T\Delta S$,the reaction becomes spontaneous when $\Delta H - T\Delta S < 0$,or $T > \frac{\Delta H}{\Delta S}$.
First,calculate $\Delta H^{\circ}_{rxn}$:
$\Delta H^{\circ}_{rxn} = [\Delta H^{\circ}_{f}(Fe) + \Delta H^{\circ}_{f}(CO)] - [\Delta H^{\circ}_{f}(FeO) + \Delta H^{\circ}_{f}(C_{(\text{graphite})})]$
$= [0 + (-110.5)] - [-266.3 + 0] = 155.8 \ \text{kJ mol}^{-1} = 155800 \ \text{J mol}^{-1}$.
Next,calculate $\Delta S^{\circ}_{rxn}$:
$\Delta S^{\circ}_{rxn} = [S^{\circ}(Fe) + S^{\circ}(CO)] - [S^{\circ}(FeO) + S^{\circ}(C_{(\text{graphite})})]$
$= [27.28 + 197.6] - [57.49 + 5.74] = 224.88 - 63.23 = 161.65 \ \text{J mol}^{-1} \text{K}^{-1}$.
The minimum temperature $T$ is given by:
$T = \frac{\Delta H^{\circ}_{rxn}}{\Delta S^{\circ}_{rxn}} = \frac{155800 \ \text{J mol}^{-1}}{161.65 \ \text{J mol}^{-1} \text{K}^{-1}} \approx 963.81 \ \text{K}$.
Rounding to the nearest integer,we get $964 \ \text{K}$.

(C) The Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T \Delta S$.
Given: $\Delta H = -57.8 \ kJ \ mol^{-1}$,$\Delta S = -176.0 \ J \ K^{-1} \ mol^{-1} = -0.176 \ kJ \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.
Substituting the values: $\Delta G = -57.8 - (298 \times -0.176)$.
$\Delta G = -57.8 + 52.448 = -5.352 \ kJ \ mol^{-1}$.
The magnitude of $\Delta G$ is $|-5.352| = 5.352 \ kJ \ mol^{-1}$.
The nearest integer value is $5$.

(A) Given: $\Delta G = -49.4 \ kJ \ mol^{-1} = -49400 \ J \ mol^{-1}$,$\Delta H = 51.4 \ kJ \ mol^{-1} = 51400 \ J \ mol^{-1}$,$T = 300 \ K$.
Using the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T \Delta S$.
Rearranging for $\Delta S$: $\Delta S = \frac{\Delta H - \Delta G}{T}$.
Substituting the values: $\Delta S = \frac{51400 - (-49400)}{300} \ J \ K^{-1} \ mol^{-1}$.
$\Delta S = \frac{100800}{300} \ J \ K^{-1} \ mol^{-1} = 336 \ J \ K^{-1} \ mol^{-1}$.

(A) In thermodynamics,the change in Gibbs free energy $( \Delta G )$ under isothermal and reversible conditions is equal to the maximum non-expansion work (or useful work) done by the system. Mathematically,$ \Delta G = W_{\text{non-expansion}} $.

(B) At equilibrium,the Gibbs free energy change $\Delta G$ is equal to $0$.
Using the relation $\Delta G = \Delta H - T \Delta S = 0$,we get $T = \frac{\Delta H}{\Delta S}$.
Given $\Delta H = -165 \ kJ \ mol^{-1} = -165000 \ J \ mol^{-1}$ and $\Delta S = -550 \ J K^{-1} mol^{-1}$.
Substituting the values: $T = \frac{-165000 \ J \ mol^{-1}}{-550 \ J K^{-1} mol^{-1}} = 300 \ K$.
Therefore,the temperature at which the reaction attains equilibrium is $300 \ K$.

(A) The combustion reaction is: $CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}$
Step $1$: Calculate the change in entropy $(\Delta S)$:
$\Delta S = S_{CO_2} - (S_{CO} + \frac{1}{2} S_{O_2})$
$\Delta S = 213.7 - (197.7 + \frac{1}{2} \times 205.1) \, J \, K^{-1} \, mol^{-1}$
$\Delta S = 213.7 - (197.7 + 102.55) = -86.55 \, J \, K^{-1} \, mol^{-1} = -0.08655 \, kJ \, K^{-1} \, mol^{-1}$
Step $2$: Calculate the change in Gibbs free energy $(\Delta G)$:
$\Delta G = \Delta H - T \Delta S$
$\Delta G = -283.0 - (298 \times -0.08655)$
$\Delta G = -283.0 + 25.79 = -257.21 \, kJ \, mol^{-1}$
Step $3$: Maximum work $(w_{max})$ is equal to $-\Delta G$:
$w_{max} = -(-257.21) \approx 257 \, kJ \, mol^{-1}$

(A) At equilibrium,the change in Gibbs free energy is $\Delta G^{\circ} = 0 \, kJ \, mol^{-1}$.
Using the relation $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$,at equilibrium,we have $0 = \Delta H^{\circ} - T \Delta S^{\circ}$.
Therefore,$T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$.
Given $\Delta H^{\circ} = 7.5 \, kJ \, mol^{-1} = 7500 \, J \, mol^{-1}$ and $\Delta S^{\circ} = 25 \, J \, K^{-1} \, mol^{-1}$.
$T = \frac{7500 \, J \, mol^{-1}}{25 \, J \, K^{-1} \, mol^{-1}} = 300 \, K$.
Thus,the values are $0 \, kJ \, mol^{-1}$ and $300 \, K$.

(B) process is spontaneous if $\Delta G < 0$ and non-spontaneous if $\Delta G > 0$. The Gibbs free energy change is given by $\Delta G = \Delta H - T \Delta S$.
For process $A$: $\Delta G = -25000 - (300 \times -80) = -25000 + 24000 = -1000 \ J \ mol^{-1}$ (Spontaneous).
For process $B$: $\Delta G = -22000 - (300 \times 40) = -22000 - 12000 = -34000 \ J \ mol^{-1}$ (Spontaneous).
For process $C$: $\Delta G = 25000 - (300 \times -50) = 25000 + 15000 = +40000 \ J \ mol^{-1}$ (Non-spontaneous).
For process $D$: $\Delta G = 22000 - (300 \times 20) = 22000 - 6000 = +16000 \ J \ mol^{-1}$ (Non-spontaneous).
Thus,processes $C$ and $D$ are non-spontaneous. The total number of non-spontaneous processes is $2$.

(D) For a spontaneous process,the slope of the $G$ vs. extent of reaction graph is negative,i.e.,$dG/d\xi < 0$.
For a process at equilibrium,the slope is zero,i.e.,$dG/d\xi = 0$.
For a non-spontaneous process,the slope is positive,i.e.,$dG/d\xi > 0$.
At point $(a)$,the slope is negative,so the reaction is spontaneous.
At point $(b)$,the slope is zero,so the reaction is at equilibrium.
At point $(c)$,the slope is positive,so the reaction is non-spontaneous.
Evaluating the statements:
$A$. Reaction is spontaneous at $(a)$ and $(b)$: False (at $(b)$ it is at equilibrium).
$B$. Reaction is at equilibrium at point $(b)$ and non-spontaneous at point $(c)$: True.
$C$. Reaction is spontaneous at $(a)$ and non-spontaneous at $(c)$: True.
$D$. Reaction is non-spontaneous at $(a)$ and $(b)$: False.
Therefore,there are $2$ true statements ($B$ and $C$).

(B) For a spontaneous reaction,$\Delta G < 0$ (negative).
For a non-spontaneous reaction,$\Delta G > 0$ (positive).
For a reversible reaction at equilibrium,$\Delta G = 0$.
Therefore,the statement '$\Delta G$ is positive for a spontaneous reaction' is incorrect.

(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative. $\Delta G = \Delta H - T\Delta S < 0$.
At equilibrium,$\Delta G = 0$,so $T = \frac{\Delta H}{\Delta S}$.
Given $\Delta H = 400 \ kJ \ mol^{-1}$ and $\Delta S = 0.2 \ kJ \ mol^{-1} \ K^{-1}$.
$T = \frac{400}{0.2} = 2000 \ K$.
Since $\Delta H$ and $\Delta S$ are both positive,the reaction becomes spontaneous at temperatures higher than the equilibrium temperature.
Therefore,the reaction will become spontaneous above $2000 \ K$.

(B) At equilibrium,the change in Gibbs free energy is $\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T \Delta S$,at equilibrium we have $\Delta H_{vap} = T \Delta S_{vap}$.
Given $\Delta H_{vap} = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$ and $\Delta S_{vap} = 75 \ J \ mol^{-1} \ K^{-1}$.
Substituting the values: $30000 \ J \ mol^{-1} = T \times 75 \ J \ mol^{-1} \ K^{-1}$.
$T = \frac{30000}{75} \ K = 400 \ K$.

(C) At equilibrium,the change in Gibbs free energy is given by $\Delta G = \Delta G^0 + P \Delta V = 0$.
Therefore,$P \Delta V = -\Delta G^0$.
Here,$\Delta G^0 = \Delta_f G^0[C(\text{diamond})] - \Delta_f G^0[C(\text{graphite})] = 2.9 \ kJ \ mol^{-1} - 0 = 2.9 \times 10^3 \ J \ mol^{-1}$.
The change in volume is $\Delta V = V_{\text{diamond}} - V_{\text{graphite}} = -2 \times 10^{-6} \ m^3 \ mol^{-1}$.
Substituting these values into $P \Delta V = -\Delta G^0$:
$P \times (-2 \times 10^{-6} \ m^3 \ mol^{-1}) = -2.9 \times 10^3 \ J \ mol^{-1}$.
$P = \frac{2.9 \times 10^3}{2 \times 10^{-6}} \ Pa = 1.45 \times 10^9 \ Pa$.
Since $1 \ bar = 10^5 \ Pa$,$P = \frac{1.45 \times 10^9}{10^5} \ bar = 1.45 \times 10^4 \ bar = 14500 \ bar$.

(A) The reaction is $N_2O_4(g) \rightarrow 2NO_2(g)$.
Given: $\Delta H^{\circ} = 55.0 \ kJ \ mol^{-1} = 55000 \ J \ mol^{-1}$.
Given: $\Delta S^{\circ} = 175.0 \ J \ K^{-1} \ mol^{-1}$.
Temperature $T = 25 + 273 = 298 \ K$.
The Gibbs free energy change is given by the formula: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.
Substituting the values: $\Delta G^{\circ} = 55000 \ J \ mol^{-1} - (298 \ K \times 175.0 \ J \ K^{-1} \ mol^{-1})$.
$\Delta G^{\circ} = 55000 - 52150 = 2850 \ J \ mol^{-1}$.

(D) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$.
For a reaction to be spontaneous,$\Delta G < 0$.
$1$. If $\Delta H > 0$ and $\Delta S < 0$,then $\Delta G$ is always positive,so the reaction is non-spontaneous at all temperatures ($A$ is correct).
$2$. If $\Delta H > 0$ and $\Delta S > 0$,then $\Delta G < 0$ only at high temperatures ($B$ is incorrect).
$3$. If $\Delta H < 0$ and $\Delta S < 0$,then $\Delta G < 0$ only at low temperatures ($C$ is correct).
$4$. If $\Delta H < 0$ and $\Delta S > 0$,then $\Delta G$ is always negative,so the reaction is spontaneous at all temperatures ($D$ is correct).
Thus,conditions $A, C,$ and $D$ are correct. However,based on the provided options,the combination $A$ and $C$ is the most appropriate match.

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,where $\Delta G = \Delta H - T \Delta S$.
Given that the reaction is endothermic,$\Delta H > 0$ (positive).
At low temperature (freezing point of water,$273 \ K$),the reaction is non-spontaneous,meaning $\Delta G > 0$. This implies $\Delta H > T \Delta S$.
At high temperature (boiling point of water,$373 \ K$),the reaction is spontaneous,meaning $\Delta G < 0$. This implies $\Delta H < T \Delta S$.
For $\Delta G$ to become negative as temperature $T$ increases,$\Delta S$ must be positive. Thus,both $\Delta H$ and $\Delta S$ are positive.

(D) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$.
$A$ reaction is spontaneous if $\Delta G < 0$.
If $\Delta H$ is $+ve$ (endothermic) and $\Delta S$ is $-ve$ (decrease in entropy),then $\Delta G = (+ve) - T(-ve) = (+ve) + T(+ve)$,which results in a positive $\Delta G$ at all temperatures.
Therefore,the reaction is non-spontaneous (impossible) under these conditions.

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,i.e.,$\Delta G < 0$.
Given the equation $\Delta G = \Delta H - T\Delta S < 0$.
Convert $\Delta H$ to Joules: $\Delta H = -84.2 \ kJ = -84200 \ J$.
Substituting the values: $-84200 \ J - T(-200 \ J \ K^{-1}) < 0$.
$-84200 + 200T < 0$.
$200T < 84200$.
$T < \frac{84200}{200} \ K$.
$T < 421 \ K$.
Thus,the highest temperature at which the reaction proceeds in the forward direction is $421 \ K$.

(A) For an isothermal reversible compression of an ideal gas,the work done $(w)$ is given by the formula: $w = -2.303 \ nRT \log\left(\frac{V_f}{V_i}\right)$.
Given: $n = 1 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$V_i = 12 \ dm^3$,$V_f = 6 \ dm^3$.
Substituting the values: $w = -2.303 \times 1 \times 8.314 \times 300 \times \log\left(\frac{6}{12}\right)$.
$w = -2.303 \times 8.314 \times 300 \times \log(0.5)$.
Since $\log(0.5) = -0.3010$,we have: $w = -2.303 \times 8.314 \times 300 \times (-0.3010)$.
$w \approx 1728.8 \ J = 1.729 \ kJ$.

(A) The work done in a chemical reaction is given by the formula $W = -\Delta n_g RT$.
First,calculate the change in the number of moles of gaseous species,$\Delta n_g = n_{p(g)} - n_{r(g)}$.
For the reaction $4 SO_{2(g)} + 2 O_{2(g)} \rightarrow 4 SO_{3(g)}$,the number of moles of gaseous products is $4$ and the number of moles of gaseous reactants is $4 + 2 = 6$.
So,$\Delta n_g = 4 - 6 = -2 \ mol$.
The temperature $T = 27^{\circ} C = 27 + 273 = 300 \ K$.
The gas constant $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Now,substitute the values into the formula: $W = -(-2 \ mol) \times (8.314 \ J \ K^{-1} \ mol^{-1}) \times (300 \ K)$.
$W = 2 \times 8.314 \times 300 = 4988.4 \ J$.

(A) For an isothermal reversible compression,the work done $w$ is given by the formula: $w = -nRT \ln\left(\frac{P_2}{P_1}\right)$.
Given values: $n = 1 \ mol$,$T = 300 \ K$,$P_1 = x \ bar$,$P_2 = 2x \ bar$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting these values into the formula:
$w = -(1 \ mol) \times (8.314 \ J \ K^{-1} \ mol^{-1}) \times (300 \ K) \times \ln\left(\frac{2x}{x}\right)$.
$w = -8.314 \times 300 \times \ln(2)$.
Using $\ln(2) \approx 0.693$:
$w = -2494.2 \times 0.693 \approx -1728.5 \ J$.
Converting to $kJ$: $w \approx -1.729 \ kJ$.
The magnitude of work done on the system is $1.729 \ kJ$.