Free energy and Work Questions in English

(C) Given: $V_{1} = 4 \ dm^{3}$,$V_{2} = 6 \ dm^{3}$,$P_{ex} = 3 \ atm$.
The formula for work done against constant external pressure is $W = -P_{ex} \Delta V$.
$\Delta V = V_{2} - V_{1} = 6 \ dm^{3} - 4 \ dm^{3} = 2 \ dm^{3} = 2 \ L$.
$W = -3 \ atm \times 2 \ L = -6 \ L \cdot atm$.
Since $1 \ L \cdot atm = 101.3 \ J$,we have $W = -6 \times 101.3 \ J = -607.8 \ J$.

(B) 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 the reaction to be spontaneous at high temperature,the term $T\Delta S$ must be large and positive to make $\Delta G$ negative,even if $\Delta H$ is positive.
This requires $\Delta S > 0$ and $\Delta H > 0$.
Thus,when $\Delta H > 0$ and $\Delta S > 0$,the reaction becomes spontaneous at high temperatures where $T\Delta S > \Delta H$.

(A) For a reaction to be spontaneous,the Gibbs free energy change must be negative,i.e.,$\Delta G = \Delta H - T\Delta S < 0$.
Given that the reaction is spontaneous below equilibrium temperature $(T < T_{eq})$,we analyze the condition $\Delta H - T\Delta S < 0$.
If $\Delta H < 0$ and $\Delta S < 0$,then $\Delta G = \Delta H - T\Delta S$.
For $\Delta G < 0$,we require $|\Delta H| > |T\Delta S|$,which implies $T < \frac{\Delta H}{\Delta S}$.
Since both $\Delta H$ and $\Delta S$ are negative,the ratio $\frac{\Delta H}{\Delta S}$ is positive,representing the equilibrium temperature.
Thus,the reaction is spontaneous when $T$ is less than this equilibrium temperature.

(A) The relationship between Gibbs free energy,enthalpy,and entropy is given by the equation: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.
Given values are: $\Delta G^{\circ} = 28000 \ J$,$\Delta S^{\circ} = 120 \ J \ K^{-1}$,and $T = 298 \ K$.
Rearranging the formula to solve for $\Delta H^{\circ}$: $\Delta H^{\circ} = \Delta G^{\circ} + T\Delta S^{\circ}$.
Substituting the values: $\Delta H^{\circ} = 28000 \ J + (298 \ K \times 120 \ J \ K^{-1})$.
$\Delta H^{\circ} = 28000 \ J + 35760 \ J = 63760 \ J$.
Converting to $kJ$: $\Delta H^{\circ} = 63.76 \ kJ$.

(C) The Gibbs free energy equation is given by $\Delta G = \Delta H - T \Delta S$.
For $\Delta G = 0$,the equation becomes $\Delta H = T \Delta S$.
Given: $\Delta H = -210 \ kJ = -210000 \ J$ and $\Delta S = -150 \ J \ K^{-1}$.
Substituting the values: $-210000 \ J = T \times (-150 \ J \ K^{-1})$.
$T = \frac{-210000}{-150} \ K = 1400 \ K$.
Therefore,the temperature is $1400 \ K$.

(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 non-spontaneous,$\Delta G$ must be greater than $0$ $(\Delta G > 0)$.
If $\Delta H > 0$ (endothermic) and $\Delta S < 0$ (decrease in entropy),then $\Delta G = (\text{positive value}) - T(\text{negative value}) = \text{positive value} + T(\text{positive value})$.
Since $T$ is always positive in Kelvin,$\Delta G$ will always be positive regardless of the temperature.
Therefore,the reaction is non-spontaneous at all temperatures when $\Delta H > 0$ and $\Delta S < 0$.

(C) The spontaneity of a reaction is determined by the Gibbs free energy change,given by the equation: $\Delta G = \Delta H - T\Delta S$.
For a reaction to be spontaneous at all temperatures,$\Delta G$ must be negative $(\Delta G < 0)$ for all values of $T$.
If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in entropy),then $\Delta G$ will always be negative regardless of the temperature $T$,because both terms ($\Delta H$ and $-T\Delta S$) will contribute to a negative value.
Therefore,the correct condition is $\Delta H < 0$ and $\Delta S > 0$.

(A) The relationship between Gibbs free energy change $(\Delta G)$,enthalpy change $(\Delta H)$,and entropy change $(\Delta S)$ is given by the equation: $\Delta G = \Delta H - T \Delta S$.
For $\Delta G = 0$,the equation becomes: $0 = \Delta H - T \Delta S$,which implies $T = \frac{\Delta H}{\Delta S}$.
Given: $\Delta H = -225 \ kJ \ mol^{-1} = -225000 \ J \ mol^{-1}$ and $\Delta S = -150 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $T = \frac{-225000 \ J \ mol^{-1}}{-150 \ J \ K^{-1} \ mol^{-1}} = 1500 \ K$.
Therefore,the temperature at which $\Delta G$ is zero is $1500 \ K$.

(B) The temperature in Kelvin is $T = 1000 + 273 = 1273 \ K$.
The formula for Gibbs energy change is $\Delta G = \Delta H - T \Delta S$.
Substituting the given values: $\Delta G = 31400 \ J - (1273 \ K \times 32 \ J \ K^{-1})$.
$\Delta G = 31400 \ J - 40736 \ J$.
$\Delta G = -9336 \ J$.

(D) Given: $\Delta H = 57.44 \ kJ$,$\Delta S = 176 \ J \ K^{-1} \ mol^{-1} = 0.176 \ kJ \ K^{-1} \ mol^{-1}$,and $T = 300 \ K$.
The Gibbs free energy change is given by the formula: $\Delta G = \Delta H - T \Delta S$.
Substituting the values: $\Delta G = 57.44 \ kJ - (300 \ K \times 0.176 \ kJ \ K^{-1} \ mol^{-1})$.
$\Delta G = 57.44 \ kJ - 52.8 \ kJ$.
$\Delta G = 4.64 \ kJ$.

(D) Given:
$\Delta H = 7 \ kJ$
$\Delta S = 24.8 \ J \ K^{-1} = 24.8 \times 10^{-3} \ kJ \ K^{-1}$
$T = 300 \ K$
The Gibbs free energy change is given by the formula:
$\Delta G = \Delta H - T \Delta S$
Substituting the values:
$\Delta G = 7 \ kJ - (300 \ K \times 24.8 \times 10^{-3} \ kJ \ K^{-1})$
$\Delta G = 7 \ kJ - 7.44 \ kJ$
$\Delta G = -0.44 \ kJ$

(C) The condition for spontaneity is given by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For the reaction to be at equilibrium,$\Delta G = 0$,which implies $T = \frac{\Delta H}{\Delta S}$.
Given: $\Delta H = -30 \ kJ = -30000 \ J$ and $\Delta S = -45 \ J \ K^{-1}$.
Substituting the values: $T = \frac{-30000 \ J}{-45 \ J \ K^{-1}} = 666.67 \ K$.
Since both $\Delta H$ and $\Delta S$ are negative,the reaction is spontaneous at temperatures below $666.67 \ K$ and becomes non-spontaneous at temperatures above $666.67 \ K$.

(C) For a reaction to be spontaneous,the Gibbs free energy change must be negative,i.e.,$\Delta G < 0$.
According to the Gibbs-Helmholtz equation,$\Delta G = \Delta H - T\Delta S$.
If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in entropy),then $\Delta G$ will always be negative at all temperatures because $\Delta H$ is negative and $-T\Delta S$ is also negative.
Therefore,the condition $\Delta S > 0, \Delta H < 0, \Delta G < 0$ represents a spontaneous reaction at all temperatures.

(B) 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 at all values of $T$.
If $\Delta H$ is negative (exothermic) and $\Delta S$ is positive (increase in entropy),then $\Delta G = (\text{negative}) - T(\text{positive})$.
Since $T$ is always positive in Kelvin,$-T\Delta S$ will always be negative.
Thus,$\Delta G$ will always be negative regardless of the temperature.

(B) For a process to be spontaneous at constant temperature and pressure,the change in Gibbs free energy must be negative.
Therefore,the criterion is $\Delta G < 0$.

(B) For a reaction where $\Delta H < 0$ and $\Delta S < 0$,the spontaneity is governed by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For the reaction to be spontaneous,$\Delta G$ must be negative.
Since $\Delta S$ is negative,the term $-T\Delta S$ is positive.
As temperature $(T)$ decreases,the magnitude of the positive term $-T\Delta S$ decreases,making $\Delta G$ more negative.
Thus,the reaction becomes more thermodynamically favorable at lower temperatures.
In the context of this question,the term "rate" refers to the thermodynamic favorability or the extent of the reaction.

(C) The conversion of oxygen into ozone is an endothermic process $(\Delta H = +ve)$ and proceeds with a decrease in entropy $(\Delta S = -ve)$.
The reaction is $3O_{2(g)} \rightleftharpoons 2O_{3(g)}$.
According to the Gibbs free energy equation,$\Delta G = \Delta H - T\Delta S$.
Since $\Delta H$ is positive and $\Delta S$ is negative,the term $-T\Delta S$ becomes positive.
Thus,$\Delta G$ is always positive at all temperatures.
Therefore,the reaction is non-spontaneous at all temperatures.

(B) According to Gibbs' energy equation,$ \Delta G = \Delta H - T \Delta S $.
For a process to be spontaneous,$ \Delta G $ must be negative.
When $ \Delta H $ is negative (exothermic) and $ \Delta S $ is positive (increase in entropy),the term $ -T \Delta S $ becomes negative.
Thus,$ \Delta G = (\text{negative}) - (\text{positive}) = \text{negative}$.
Therefore,the process is spontaneous at all temperatures when $ \Delta H < 0 $ and $ \Delta S > 0 $.

(C) For an isothermal,reversible expansion of an ideal gas,the work done is given by the formula: $W = -nRT \ln(P_1/P_2)$.
Given:
Mass of $H_2$ gas = $10 \ g$.
Molar mass of $H_2$ = $2 \ g \ mol^{-1}$.
Number of moles $(n) = 10 \ g / 2 \ g \ mol^{-1} = 5 \ mol$.
Temperature $(T) = 273 \ K$.
Initial pressure $(P_1) = 10 \ atm$.
Final pressure $(P_2) = 1 \ atm$.
Gas constant $(R) = 8.3 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values:
$W = -5 \ mol \times 8.3 \ J \ K^{-1} \ mol^{-1} \times 273 \ K \times \ln(10/1)$.
$W = -5 \times 8.3 \times 273 \times 2.303 \times \log(10)$.
$W = -11331.45 \times 2.303 \times 1 \ J$.
$W \approx -26096 \ J = -26.096 \ kJ$.
Since the work is done by the system,it is negative. Thus,the correct option is $C$.

(A) The process of vaporization is non-spontaneous below the boiling point at $1 \ atm$ pressure.
For any process,the spontaneity is determined by the Gibbs free energy change,$\Delta G = \Delta H - T \Delta S$.
At the boiling point $(T_{b} = 373 \ K)$,the system is at equilibrium,so $\Delta G = 0$,which implies $\Delta S = \Delta H / T_{b}$.
For vaporization,$\Delta H > 0$ and $\Delta S > 0$.
At $T = 75^{\circ} C = 348 \ K$,since $T < T_{b}$,the term $T \Delta S$ is less than $\Delta H$.
Therefore,$\Delta G = \Delta H - T \Delta S > 0$.
Since $\Delta G > 0$,the process is non-spontaneous at $75^{\circ} C$.

(B) The relationship between Gibbs free energy,enthalpy,and entropy is given by the equation: $\Delta G^{\ominus} = \Delta H^{\ominus} - T \Delta S^{\ominus}$.
Given values are: $\Delta G^{\ominus} = -9 \ kJ = -9000 \ J$,$\Delta H^{\ominus} = -24 \ kJ = -24000 \ J$,and $T = 25 + 273 = 298 \ K$.
Substituting these values into the equation: $-9000 = -24000 - (298 \times \Delta S^{\ominus})$.
Rearranging the terms: $298 \times \Delta S^{\ominus} = -24000 + 9000 = -15000$.
Solving for $\Delta S^{\ominus}$: $\Delta S^{\ominus} = -15000 / 298 \approx -50.33 \ J \ K^{-1}$.

(D) The Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T \Delta S$.
Given: $\Delta H = -145 \ kJ \ mol^{-1}$,$\Delta S = -650 \ J \ K^{-1} \ mol^{-1} = -0.650 \ kJ \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.
Substituting the values: $\Delta G = -145 \ kJ \ mol^{-1} - (298 \ K \times -0.650 \ kJ \ K^{-1} \ mol^{-1})$.
$\Delta G = -145 + 193.7 = +48.7 \ kJ \ mol^{-1}$.
Since $\Delta G > 0$,the reaction is non-spontaneous.

(C) 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,$\Delta G < 0$.
For a reaction to be non-spontaneous,$\Delta G > 0$.
Given that the reaction is non-spontaneous at $298 \ K$ $(\Delta G > 0)$ and spontaneous at $350 \ K$ $(\Delta G < 0)$,the reaction must be endothermic $(\Delta H > 0)$ and have a positive entropy change $(\Delta S > 0)$.
At lower temperatures,the $T \Delta S$ term is smaller than $\Delta H$,making $\Delta G$ positive.
At higher temperatures,the $T \Delta S$ term exceeds $\Delta H$,making $\Delta G$ negative.
Thus,the conditions are $\Delta H > 0$ and $\Delta S > 0$.

(B) The spontaneity of a reaction is determined by the Gibbs free energy change $(\Delta G)$,which is calculated using the formula: $\Delta G = \Delta H - T \Delta S$.
Given: $\Delta H = -2.5 \times 10^3 \ cal$,$\Delta S = 7.4 \ cal \ K^{-1}$,and $T = 298 \ K$.
Substituting the values: $\Delta G = (-2.5 \times 10^3) - (298 \times 7.4)$.
$\Delta G = -2500 - 2205.2 = -4705.2 \ cal$.
Since $\Delta G < 0$,the reaction is spontaneous.

(B) The spontaneity of a reaction is determined by the Gibbs free energy change,given by the equation $\Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus}$.
For a reaction to be spontaneous,$\Delta_r G^{\ominus} < 0$.
$(A)$ If $\Delta_r H^{\ominus} < 0$ and $\Delta_r S^{\ominus} > 0$,then $\Delta_r G^{\ominus} = (-H) - T(+S) = (-H) - (T \times S)$,which is always negative at all temperatures. This is correct.
$(B)$ If $\Delta_r H^{\ominus} < 0$ and $\Delta_r S^{\ominus} < 0$,then $\Delta_r G^{\ominus} = (-H) - T(-S) = -H + TS$. For spontaneity,$-H + TS < 0$,which means $TS < H$ or $T < H/S$. Thus,it is spontaneous at low temperatures and non-spontaneous at high temperatures. Option $(B)$ states it is non-spontaneous at low temperatures,which is incorrect.
$(C)$ If $\Delta_r H^{\ominus} > 0$ and $\Delta_r S^{\ominus} > 0$,then $\Delta_r G^{\ominus} = (+H) - T(+S)$. For spontaneity,$T > H/S$. Thus,it is non-spontaneous at low temperatures. This is correct.
$(D)$ If $\Delta_r H^{\ominus} < 0$ and $\Delta_r S^{\ominus} < 0$,it is spontaneous at low temperatures (as derived in $B$). This is correct.
Therefore,the incorrect option is $(B)$.

(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta_r G^{\circ}$ must be negative.
The relationship is given by the equation: $\Delta_r G^{\circ} = \Delta_r H^{\circ} - T \Delta_r S^{\circ}$.
For the reaction to be spontaneous at all temperatures $(T)$,the term $\Delta_r G^{\circ}$ must remain negative regardless of the value of $T$.
This occurs when $\Delta_r H^{\circ}$ is negative (exothermic) and $\Delta_r S^{\circ}$ is positive (increase in entropy).
Thus,the correct signs are negative and positive.

(B) The spontaneity of a reaction is determined by the Gibbs free energy change,given by the equation: $\Delta G = \Delta H - T \Delta S$.
For a reaction to be spontaneous,$\Delta G$ must be negative $(\Delta G < 0)$.
If $\Delta H > 0$ (endothermic) and $\Delta S > 0$ (increase in entropy),the reaction becomes spontaneous only when the term $T \Delta S$ is larger than $\Delta H$,which occurs at high temperatures.

Condition	Spontaneity
$\Delta H > 0, \Delta S > 0$	Spontaneous at high $T$
$\Delta H < 0, \Delta S < 0$	Spontaneous at low $T$
$\Delta H < 0, \Delta S > 0$	Spontaneous at all $T$
$\Delta H > 0, \Delta S < 0$	Non-spontaneous at all $T$

(B) For a reaction to be spontaneous,$\Delta G$ must be negative $(\Delta G < 0)$.
According to the Gibbs-Helmholtz equation,$\Delta G = \Delta H - T\Delta S$.
If $\Delta H < 0$ and $\Delta S < 0$,then $\Delta G = \Delta H - T\Delta S$ will be negative only when $|\Delta H| > |T\Delta S|$,which occurs at low temperatures.
At high temperatures,$|T\Delta S| > |\Delta H|$,making $\Delta G$ positive (non-spontaneous).
Therefore,$\Delta H < 0$ and $\Delta S < 0$ at high temperature is not suitable for a spontaneous reaction.

(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative. The relationship is given by $\Delta G = \Delta H - T \Delta S$.
Given $\Delta H > 0$ (endothermic) and $\Delta S > 0$ (increase in entropy).
For $\Delta G < 0$,the magnitude of $T \Delta S$ must be greater than $\Delta H$.
This condition is satisfied at high temperatures.
Therefore,the reaction is spontaneous at high temperatures.
The correct option is $B$.

(B) reaction is spontaneous when the change in Gibbs free energy $( \Delta G )$ is negative.
For a spontaneous process,$ \Delta G < 0 $ at constant temperature and pressure.
If $ \Delta G = 0 $,the reaction is at equilibrium.
If $ \Delta G > 0 $,the reaction is non-spontaneous.

(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative. $\Delta G = \Delta H - T \Delta S$.
For the reaction to be in equilibrium,$\Delta G = 0$,so $T = \frac{\Delta H}{\Delta S}$.
Given: $\Delta H = -41.2 \ kJ = -41200 \ J$ and $\Delta S = -42.4 \ J \ K^{-1}$.
Calculating the equilibrium temperature: $T = \frac{-41200 \ J}{-42.4 \ J \ K^{-1}} \approx 971.7 \ K$.
For the reaction to proceed in the reverse direction,the reverse reaction must be spontaneous,which means $\Delta G_{reverse} < 0$,or $\Delta G_{forward} > 0$.
Since $\Delta H$ and $\Delta S$ are both negative,the forward reaction is spontaneous at temperatures below the equilibrium temperature $(T < 971.7 \ K)$.
Therefore,the reaction will proceed in the reverse direction at temperatures above $971.7 \ K$.

(B) The Gibbs function $G$ is defined as $G = H - T S$.
Taking the differential of both sides,we get $d G = d H - T d S - S d T$ $(i)$.
From the definition of enthalpy,$H = U + p V$,so $d H = d U + p d V + V d p$.
Substituting the first law of thermodynamics $d U = T d S - p d V$ into the expression for $d H$,we get $d H = (T d S - p d V) + p d V + V d p = T d S + V d p$.
Now,substitute $d H = T d S + V d p$ into equation $(i)$:
$d G = (T d S + V d p) - T d S - S d T$
$d G = V d p - S d T$.

(C) For a reaction to be spontaneous,the Gibbs free energy change must be negative,i.e.,$\Delta_r G^{\circ} < 0$.
The relationship is given by: $\Delta_r G^{\circ} = \Delta_r H^{\circ} - T \Delta_r S^{\circ}$.
Analyzing the table:
$1$. For row $B$: $\Delta_r H^{\circ} = (+)$,$\Delta_r S^{\circ} = (+)$. $\Delta_r G^{\circ}$ is negative when $T \Delta_r S^{\circ} > \Delta_r H^{\circ}$,which occurs at high temperatures. Thus,$B$ is correct.
$2$. For row $C$: $\Delta_r H^{\circ} = (-)$,$\Delta_r S^{\circ} = (-)$. $\Delta_r G^{\circ}$ is negative when $|\Delta_r H^{\circ}| > |T \Delta_r S^{\circ}|$,which occurs at low temperatures. Thus,$C$ is correct.
Rows $A$ and $D$ show positive $\Delta_r G^{\circ}$ values,which correspond to non-spontaneous processes under the conditions listed.
Therefore,options $B$ and $C$ are correct.

(B) Given: $\Delta H = +30.0 \ kJ \ mol^{-1}$,$\Delta S = 0.06 \ kJ \ K^{-1} \ mol^{-1}$.
For the reaction to be at equilibrium,$\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T \Delta S$,we set $\Delta G = 0$:
$0 = 30.0 - T \times 0.06$
$T = \frac{30.0}{0.06} = 500 \ K$.
Converting to Celsius: $T(^{\circ} C) = 500 - 273 = 227^{\circ} C$.
Since $\Delta H$ is positive (endothermic) and $\Delta S$ is positive,the reaction is spontaneous only at temperatures above $500 \ K$ (or $227^{\circ} C$).
Therefore,below $227^{\circ} C$,the reaction is non-spontaneous.

(A) The correct statement is $A$. $\Delta G = \Delta G^{\circ}$ when the system is at the standard state (where all reactants and products are at $1 \ M$ concentration or $1 \ bar$ pressure).
Analysis of other options:
$B$. At equilibrium,$\Delta G = 0$,not $\Delta G^{\circ} = 0$.
$C$. $\Delta G$ measures the spontaneity of a reaction,not the activation energy.
$D$. When $\Delta G > 0$,the reaction is non-spontaneous in the forward direction and will proceed in the backward direction.

(B) For a reaction,the Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T \Delta S$.
Given that $\Delta G = 0$,we have $0 = \Delta H - T \Delta S$,which implies $T = \frac{\Delta H}{\Delta S}$.
Substituting the given values: $\Delta H = -20.5 \ kJ \ mol^{-1} = -20500 \ J \ mol^{-1}$ and $\Delta S = -50.0 \ J \ K^{-1} \ mol^{-1}$.
$T = \frac{-20500 \ J \ mol^{-1}}{-50.0 \ J \ K^{-1} \ mol^{-1}} = 410 \ K$.

(C) For a reaction to be spontaneous,$\Delta G < 0$.
Since $\Delta G = \Delta H - T\Delta S$,the condition for spontaneity is $\Delta H - T\Delta S < 0$.
Given: $\Delta H = -400 \text{ kJ mol}^{-1}$ and $\Delta S = -20 \text{ kJ mol}^{-1} \text{ K}^{-1}$.
Substituting the values: $-400 - T(-20) < 0$.
$-400 + 20T < 0$.
$20T < 400$.
$T < 20 \text{ K}$.
Therefore,the reaction is spontaneous below $20 \text{ K}$.

(A) For a spontaneous process,the following criteria must be satisfied:
$1$. The change in Gibbs energy at constant temperature and pressure must be negative: $(\Delta G_{\text{system}})_{T, P} < 0$.
$2$. The total entropy change of the universe must be positive: $\Delta S_{\text{total}} = (\Delta S_{\text{system}}) + (\Delta S_{\text{surroundings}}) > 0$.
$3$. The change in internal energy at constant entropy and volume must be negative: $(\Delta U_{\text{system}})_{S, V} < 0$.
Therefore,the statement $(\Delta G_{\text{system}})_{T, P} > 0$ is incorrect.

(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative.
The relationship is given by the equation $\Delta G = \Delta H - T \Delta S$.
For $\Delta G$ to be negative at all temperatures $(T)$,the enthalpy change $\Delta H$ must be negative (exothermic) and the entropy change $\Delta S$ must be positive (increase in disorder).
Thus,the condition is $\Delta H < 0$ and $\Delta S > 0$.

(C) For a reaction to be spontaneous at constant temperature and pressure,the change in Gibbs free energy,$\Delta G$,must be negative.
The relationship is given by the equation: $\Delta G = \Delta H - T \Delta S$.
Therefore,the condition for spontaneity is $\Delta G < 0$,which implies $(\Delta H - T \Delta S) < 0$.

(B) From the Gibbs-Helmholtz equation,$\Delta G = \Delta H - T \Delta S$.
For a process to be spontaneous at constant temperature and pressure,the change in Gibbs free energy of the system must be negative,i.e.,$\Delta G < 0$.
This implies a decrease or lowering of the Gibbs free energy of the system.

(D) Given that at the transition temperature,$\Delta G^0 = 0$.
Substituting the given expression: $105 - 35 \log T = 0$.
Rearranging the equation: $35 \log T = 105$.
Dividing both sides by $35$: $\log T = 3$.
Converting from logarithmic form to exponential form: $T = 10^3 = 1000 \text{ K}$.
To convert the temperature from Kelvin to Celsius: $T(^{\circ}\text{C}) = T(\text{K}) - 273$.
$T(^{\circ}\text{C}) = 1000 - 273 = 727 \text{ }^{\circ}\text{C}$.