Free energy and Work Questions in English

(C) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For a process to be spontaneous,$\Delta G$ must be negative.
In option $(C)$,the reaction is exothermic $(\Delta H < 0)$ and there is an increase in disorder $(\Delta S > 0)$.
Substituting these into the equation: $\Delta G = (-ve) - T(+ve) = -ve$.
Since $\Delta G$ is always negative regardless of the temperature $T$,the reaction is certain to be spontaneous.

(D) For a reaction to be spontaneous,the Gibbs free energy change $ \Delta G = \Delta H - T \Delta S $ must be negative $( \Delta G < 0 )$.
If $ \Delta H $ is $ +ve $ (endothermic) and $ \Delta S $ is $ -ve $ (decrease in entropy),then $ \Delta G = (+ve) - T(-ve) = (+ve) + (T \times +ve) = +ve $.
Since $ \Delta G $ is always positive regardless of the temperature $ T $,the reaction is impossible under any conditions.

(A) The Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T\Delta S$.
Given: $\Delta H = -2.5 \times 10^3 \ cal$,$\Delta S = 7.4 \ cal \ K^{-1}$,and $T = 298 \ K$.
Calculating $\Delta G$: $\Delta G = (-2500) - (298 \times 7.4) = -2500 - 2205.2 = -4705.2 \ cal$.
Since $\Delta G < 0$,the reaction is spontaneous.

(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G = \Delta H - T\Delta S$ must be negative $(\Delta G < 0)$.
If $\Delta H > 0$ (endothermic) and $\Delta S < 0$ (decrease in entropy),then $\Delta G = (+\Delta H) - T(-\Delta S) = \Delta H + T\Delta S$.
Since both terms are positive,$\Delta G$ will always be positive regardless of the temperature $T$.
Therefore,the reaction is non-spontaneous under all conditions.

(C) The spontaneity of a process is determined by the Gibbs free energy change,given by the equation $\Delta G = \Delta H - T\Delta S$.
For a process to be non-spontaneous,$\Delta G$ must be positive $(> 0)$.
When $\Delta H > 0$ (endothermic) and $\Delta S < 0$ (decrease in entropy),the term $\Delta G = (+\Delta H) - T(-\Delta S) = \Delta H + T\Delta S$ will always be positive at all temperatures $T$.
Therefore,the condition $\Delta H > 0$ and $\Delta S < 0$ always leads to a non-spontaneous change.

(B) For a process occurring at constant temperature and pressure,the Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T\Delta S$.
At equilibrium,the process is reversible,and the change in Gibbs free energy is zero,i.e.,$\Delta G = 0$.
Substituting this into the equation,we get $0 = \Delta H - T\Delta S$,which implies $\Delta H = T\Delta S$.

(B) For a process to be spontaneous,the change in Gibbs free energy must be negative.
Therefore,$\Delta G < 0$.

(A) For a chemical reaction,the spontaneity is determined by the Gibbs free energy change,denoted as $\Delta G$.
If $\Delta G < 0$ (negative value),the reaction is spontaneous.
If $\Delta G > 0$ (positive value),the reaction is non-spontaneous.
If $\Delta G = 0$,the system is in equilibrium.
Therefore,a minus sign of the free energy change denotes that the reaction tends to proceed spontaneously.

(B) The Gibbs-Helmholtz equation describes the relationship between Gibbs free energy $(\Delta G)$,enthalpy $(\Delta H)$,temperature $(T)$,and entropy $(\Delta S)$.
It is given by the formula: $\Delta G = \Delta H - T \Delta S$.

(C) The equation $\Delta G = \Delta H - T\Delta S$ is known as the $Gibbs-Helmholtz$ equation,which relates the change in Gibbs free energy $(\Delta G)$ to the change in enthalpy $(\Delta H)$ and entropy $(\Delta S)$ at a constant temperature $(T)$.

(C) For a spontaneous process at constant pressure and temperature,the change in Gibbs free energy must be negative.
Therefore,the reaction proceeds in the direction where there is a decrease in Gibbs free energy,i.e.,$\Delta G < 0$.

(A) For an isothermal process of an ideal gas,the change in Gibbs free energy is related to pressure by the equation: $\Delta G = \int_{P_1}^{P_2} V \, dP$.
Since $PV = nRT$,we have $V = \frac{nRT}{P}$.
Substituting this into the integral: $\Delta G = \int_{P_1}^{P_2} \frac{nRT}{P} \, dP = nRT \int_{P_1}^{P_2} \frac{1}{P} \, dP$.
Evaluating the integral gives: $\Delta G = nRT \ln \left( \frac{P_2}{P_1} \right)$.
Therefore,the correct option is $A$.

(A) The condition $\Delta G = 0$ signifies that the system is in a state of thermodynamic equilibrium.
For the phase transition $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ at $1 \ atm$ and $373 \ K$ (the boiling point of water),the liquid and gaseous phases coexist in equilibrium.
Therefore,the correct statement is that $H_2O_{(l)}$ is in equilibrium with $H_2O_{(g)}$.

(D) The process of converting $1.0 \ mole$ of water into steam at $100 \ ^oC$ and $1 \ atm$ pressure represents a phase transition (boiling of water) occurring at its boiling point.
At the boiling point,the liquid and vapor phases are in equilibrium.
For any process occurring at equilibrium,the change in Gibbs free energy is $\Delta G = 0$.

(D) The standard free energy change is calculated using the Gibbs-Helmholtz equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.
Given: $\Delta H^\circ = 58.04 \, kJ = 58040 \, J$,$\Delta S^\circ = 176.7 \, J/K$,and $T = 25 + 273 = 298 \, K$.
Substituting the values: $\Delta G^\circ = 58040 \, J - (298 \, K \times 176.7 \, J/K)$.
$\Delta G^\circ = 58040 \, J - 52656.6 \, J = 5383.4 \, J$.
Converting to $kJ$: $\Delta G^\circ = 5.3834 \, kJ \approx 5.39 \, kJ$.

(D) The spontaneity of a chemical reaction at constant temperature and pressure is determined by the Gibbs free energy change,$\Delta G$.
For a spontaneous process,the change in Gibbs free energy must be negative,i.e.,$\Delta G < 0$.

(B) The Gibbs free energy change is given by the formula: $\Delta G = \Delta H - T\Delta S$.
Given: $\Delta H = -11.7 \times 10^3 \, J \, mol^{-1}$,$\Delta S = -105 \, J \, mol^{-1} \, K^{-1}$,and $T = 25 + 273 = 298 \, K$.
Substituting the values: $\Delta G = -11.7 \times 10^3 - (298 \times -105)$.
$\Delta G = -11700 + 31290 = 19590 \, J \, mol^{-1}$.
Converting to $kJ$: $\Delta G = 19590 / 1000 = 19.59 \, kJ \, mol^{-1}$.

(B) The Gibbs free energy $G$,enthalpy $H$,and entropy $S$ are related by the fundamental thermodynamic equation $G = H - TS$.
Similarly,for a process at constant temperature $T$,the change in these state functions is given by $\Delta G = \Delta H - T \Delta S$.

(C) For a reaction to be feasible or spontaneous,the change in Gibbs free energy $(\Delta G)$ must be negative.
Thus,the condition is $\Delta G < 0$.

(B) The transformation is $C_{\text{graphite}} \rightarrow C_{\text{diamond}}$.
For this process, $\Delta G = \Delta G^o + \int_{P_1}^{P_2} \Delta V \, dP = 0$ at equilibrium.
$\Delta G^o = -\Delta V \times P$ (assuming $\Delta V$ is constant).
$\Delta V = V_{\text{diamond}} - V_{\text{graphite}} = \frac{M}{\rho_{\text{diamond}}} - \frac{M}{\rho_{\text{graphite}}}$.
Given $M = 12 \, g \, mol^{-1}$, $\rho_{\text{diamond}} = 3.31 \, g \, cm^{-3}$, $\rho_{\text{graphite}} = 2.25 \, g \, cm^{-3}$.
$\Delta V = 12 \times (\frac{1}{3.31} - \frac{1}{2.25}) \, cm^3 \, mol^{-1} = 12 \times (0.3021 - 0.4444) = -1.7076 \, cm^3 \, mol^{-1} = -1.7076 \times 10^{-6} \, m^3 \, mol^{-1}$.
$P = \frac{-\Delta G^o}{\Delta V} = \frac{-1895 \, J \, mol^{-1}}{-1.7076 \times 10^{-6} \, m^3 \, mol^{-1}} \approx 1.11 \times 10^9 \, Pa$.
Given the options, the closest value is $9.92 \times 10^8 \, Pa$.

(A) The Gibbs free energy change $(\Delta G)$ for a process at constant temperature and pressure is given by the equation:
$\Delta G = \Delta H - T\Delta S$
where $\Delta H$ is the change in enthalpy,$T$ is the absolute temperature,and $\Delta S$ is the change in entropy.

(D) The standard Gibbs energy change $(\Delta G^{\circ})$ is calculated using the relation: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.
Given:
$\Delta H^{\circ} = -382.64 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = -145.6 \ J \ K^{-1} \ mol^{-1} = -0.1456 \ kJ \ K^{-1} \ mol^{-1}$
$T = 298 \ K$
Substituting the values:
$\Delta G^{\circ} = -382.64 \ kJ \ mol^{-1} - (298 \ K \times -0.1456 \ kJ \ K^{-1} \ mol^{-1})$
$\Delta G^{\circ} = -382.64 + 43.3888$
$\Delta G^{\circ} = -339.2512 \ kJ \ mol^{-1} \approx -339.3 \ kJ \ mol^{-1}$.

(A) The Gibbs free energy change is given by the formula: $\Delta G = \Delta H - T\Delta S$.
Given: $\Delta H = 31400 \, cal$,$\Delta S = 32 \, cal \, K^{-1} \, mol^{-1}$,and $T = 1000 + 273 = 1273 \, K$.
Substituting the values: $\Delta G = 31400 - (1273 \times 32)$.
$\Delta G = 31400 - 40736 = -9336 \, cal$.

(C) The reaction is endothermic because $\Delta H > 0$ $(9.08 \ kJ \ mol^{-1})$.
To determine spontaneity,we use the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For the reaction to be spontaneous,$\Delta G$ must be negative $(\Delta G < 0)$.
$\Delta G = 9080 \ J \ mol^{-1} - T(35.7 \ J \ K^{-1} \ mol^{-1})$.
Setting $\Delta G < 0$,we get $9080 < T(35.7)$,which implies $T > 254.34 \ K$.
Since the reaction can be spontaneous at temperatures above $254.34 \ K$ and is endothermic,option $(C)$ is the most appropriate description.

(A) For a reaction to be spontaneous,the change in Gibbs free energy,$\Delta G$,must be negative.
Since $\Delta G = \Delta H - T\Delta S$,the expression $(\Delta H - T\Delta S)$ must be negative.

(A) At $0 \, ^\circ C$ and $1 \, atm$ pressure,ice and liquid water are in a state of equilibrium.
For any process at equilibrium,the change in Gibbs free energy is $\Delta G = 0$.
Since the melting of water at its melting point is a reversible process at equilibrium,$\Delta G^o = 0$.

(A) The Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T\Delta S$.
Given: $\Delta H = -285.8 \ kJ \ mol^{-1}$,$\Delta S = -0.163 \ kJ \ mol^{-1} K^{-1}$,and $T = 27 + 273 = 300 \ K$.
Substituting the values: $\Delta G = (-285.8) - (300 \times -0.163)$.
$\Delta G = -285.8 + 48.9 = -236.9 \ kJ \ mol^{-1}$.

(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 non-spontaneous (opposed),$\Delta G$ must be positive $(> 0)$.
When $\Delta H = +ve$ (endothermic) and $\Delta S = -ve$ (decrease in entropy),the term $\Delta G = (+ve) - T(-ve) = (+ve) + T(+ve)$ will always be positive regardless of the temperature $T$.
Therefore,the condition $\Delta H = +ve$ and $\Delta S = -ve$ will always oppose the reaction.

(A) 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,$\Delta G = \Delta H - T\Delta S$.
For a reaction to be spontaneous at all temperatures,the enthalpy change $\Delta H$ must be negative (exothermic) and the entropy change $\Delta S$ must be positive.
Thus,the condition is $\Delta G < 0$ and $\Delta H < 0$.

(B) For a process at constant temperature and pressure,the condition for equilibrium is defined by the Gibbs free energy change,$\Delta G = 0$.
Since $\Delta G = \Delta H - T \Delta S$,setting $\Delta G = 0$ gives $\Delta H = T \Delta S$.

(A) For a process to be spontaneous,the Gibbs free energy change,$\Delta G$,must be negative $(\Delta G < 0)$.
The relationship is given by the equation: $\Delta G = \Delta H - T \Delta S$.
If $\Delta H > 0$ (endothermic) and $\Delta S > 0$ (increase in entropy),the process becomes spontaneous when $T \Delta S > \Delta H$,as this makes $\Delta G$ negative.
Therefore,the correct condition is $T \Delta S > \Delta H, \Delta H > 0, \Delta S > 0$.

(B) For a process to be spontaneous,the Gibbs free energy change,$\Delta G$,must be negative.
The relationship is given by the equation: $\Delta G = \Delta H - T \Delta S$.
Therefore,for spontaneity,the value of $(\Delta H - T \Delta S)$ must be negative.

(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be less than $0$.
Given $\Delta G = \Delta H - T\Delta S < 0$.
Substituting the values: $170,000 \ J - T(170 \ J/K) < 0$.
$170,000 < 170T$.
$T > 1000 \ K$.
Therefore,the reaction becomes spontaneous at temperatures above $1000 \ K$. Among the given options,$1110 \ K$ is the only value greater than $1000 \ K$.

(D) For a process to be spontaneous,the Gibbs free energy change must be negative,i.e.,$\Delta G < 0$.
Given $\Delta G = \Delta H - T \Delta S$,the condition for spontaneity is $\Delta H - T \Delta S < 0$.
This implies $T > \frac{\Delta H}{\Delta S}$.
Given $\Delta H = +179.1 \ kJ \ mol^{-1} = 179100 \ J \ mol^{-1}$ and $\Delta S = +160.2 \ J \ K^{-1} \ mol^{-1}$.
$T > \frac{179100}{160.2} \ K$.
$T > 1117.97 \ K$.
Therefore,the process becomes spontaneous at temperatures higher than $1118 \ K$.

(A) Given: $\Delta H = 40.63 \ kJ \ mol^{-1} = 40630 \ J \ mol^{-1}$
$\Delta S = 100 \ J \ K^{-1} \ mol^{-1}$
$T = 27 \ ^oC = 27 + 273 = 300 \ K$
The Gibbs free energy equation is $\Delta G = \Delta H - T \Delta S$.
Substituting the values: $\Delta G = 40630 - (300 \times 100)$
$\Delta G = 40630 - 30000 = 10630 \ J \ mol^{-1}$.

(C) The Gibbs free energy equation is given by $\Delta G = \Delta H - T\Delta S$.
For a spontaneous process,the condition is $\Delta G < 0$.
For an endothermic reaction,the enthalpy change is positive,i.e.,$\Delta H > 0$.
Substituting these into the equation: $\Delta G = (+\Delta H) - T\Delta S < 0$.
This implies that $T\Delta S > \Delta H$. Since $\Delta H$ is positive and $T$ is positive,$\Delta S$ must be positive $(\Delta S > 0)$ for the reaction to be spontaneous.

(C) Gibbs free energy $(G)$ is a thermodynamic potential that measures the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.
It is defined by the equation: $G = H - TS$,where $H$ is enthalpy,$T$ is absolute temperature,and $S$ is entropy.

(C) For a process to be spontaneous,the total entropy change of the universe must be positive,i.e.,$\Delta S_{total} > 0$.
At constant temperature and pressure,the spontaneity is determined by the Gibbs free energy change,$\Delta G = \Delta H - T\Delta S$.
For a spontaneous process at constant temperature and pressure,$\Delta G < 0$.
Therefore,the free energy of the system decreases.

(D) The work done in a reversible isothermal expansion is given by the formula: $W = -2.303 \ nRT \ \log \left( \frac{V_2}{V_1} \right)$.
Given values: $n = 2 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,$V_2 = 20 \ L$,$V_1 = 10 \ L$.
Substituting these values: $W = -2.303 \times 2 \times 8.314 \times 298 \times \log \left( \frac{20}{10} \right)$.
$W = -2.303 \times 2 \times 8.314 \times 298 \times \log(2)$.
$W = -2.303 \times 2 \times 8.314 \times 298 \times 0.3010 \approx -3434.9 \ J$.

(C) The work done in an expansion process is given by the formula $W = -P_{ext} \times \Delta V$.
Since the gas is expanding into a vacuum bulb,the external pressure $P_{ext}$ is equal to $0 \ atm$.
Therefore,$W = -0 \times \Delta V = 0 \ J$.

(B) According to the Gibbs free energy equation,$\Delta G = \Delta H - T \Delta S$.
For a process to be non-spontaneous,$\Delta G$ must be positive $(> 0)$.
If $\Delta H > 0$ (endothermic) and $\Delta S < 0$ (decrease in entropy),then $\Delta G = (\text{positive}) - T(\text{negative}) = (\text{positive}) + (\text{positive}) = \text{positive}$.
Therefore,the condition $\Delta H > 0$ and $\Delta S < 0$ always results in a non-spontaneous process.

(C) The change in Gibbs free energy is given by the equation: $\Delta G = \Delta H - T\Delta S$.
For the reaction to be at equilibrium,$\Delta G = 0$.
Therefore,$0 = \Delta H - T\Delta S$,which implies $T = \frac{\Delta H}{\Delta S}$.
Given $\Delta H = 30.56 \, kJ \, mol^{-1} = 30560 \, J \, mol^{-1}$ and $\Delta S = 66 \, J \, K^{-1} \, mol^{-1}$.
$T = \frac{30560}{66} \approx 463 \, K$.

(B) 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,$\Delta H$ must be negative (exothermic) and $\Delta S$ must be positive (increase in entropy/disorder).

(C) The melting of ice is a phase transition process.
At $273 \ K$ $(0 \ ^\circ C)$,ice and water are in equilibrium,so $\Delta G = 0$.
For temperatures above $273 \ K$,the melting of ice is a spontaneous process.
Since the process is spontaneous at $283 \ K$,the change in Gibbs free energy must be negative,i.e.,$\Delta G < 0$.

(B) For a spontaneous process,the Gibbs free energy of the system decreases,i.e.,$\Delta G < 0$.

(C) According to the Gibbs free energy equation,$\Delta G = \Delta H - T \Delta S$.
For a reaction to be spontaneous,$\Delta G$ must be negative.
If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in entropy),then $\Delta G$ will always be negative regardless of the temperature $T$.
Therefore,the reaction is spontaneous under these conditions.

(A) The external pressure $P_{ext} = 200 \ torr = \frac{200}{760} \ atm \approx 0.26316 \ atm$.
The change in volume $\Delta V = 3.3 \ L$.
The work done $W = -P_{ext} \times \Delta V$.
$W = -0.26316 \ atm \times 3.3 \ L = -0.8684 \ L \cdot atm$.
Since $1 \ L \cdot atm = 101.325 \ J$,
$W = -0.8684 \times 101.325 \ J \approx -87.99 \ J$.

(A) At $0^o \text{C}$ $(273 \ K)$,ice and water exist in equilibrium: $H_2O_{(s)} \rightleftharpoons H_2O_{(l)}$.
For any process at equilibrium,the change in Gibbs free energy is $\Delta G = 0$.

(D) For a spontaneous (irreversible) process occurring at constant $T$ and $P$,the criteria for spontaneity are:
$1$. The total entropy change of the universe must be positive: $(\Delta S)_{total} > 0$.
$2$. The change in Gibbs free energy must be negative: $(\Delta G)_{T,P} < 0$.
Therefore,the correct condition is $(\Delta S)_{total} > 0$ and $(\Delta G) < 0$.