Free energy and Work Questions in English

(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,where $\Delta G = \Delta H - T \Delta S$.
Since the forward reaction is spontaneous at room temperature,$\Delta H$ must be negative and $\Delta S$ must be positive for $\Delta G$ to be negative at low temperatures.
Conversely,for the reverse reaction to be spontaneous at high temperatures,the conditions must favor the reverse process,confirming $\Delta H < 0$ and $\Delta S > 0$.

(C) Given $\Delta H = 9.08 \, kJ/mol$ (positive value,so the reaction is endothermic).
For a reaction to be spontaneous,$\Delta G = \Delta H - T\Delta S$ must be negative.
At higher temperatures,$T\Delta S > \Delta H$,making $\Delta G$ negative.
Since $\Delta H > 0$,the reaction is endothermic.
At sufficiently high temperatures,the reaction becomes spontaneous.

(B) For a process at constant temperature and pressure:
$1$. If $\Delta G_{\text{system}} < 0$,the process is spontaneous.
$2$. If $\Delta G_{\text{system}} = 0$,the system is in equilibrium.
$3$. If $\Delta G_{\text{system}} > 0$,the process is non-spontaneous.

(A) 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.
Since $\Delta H > 0$ and $\Delta S > 0$,the term $T \Delta S$ must be greater than $\Delta H$ to make $\Delta G$ negative.
This is achieved by increasing the temperature $T$.

(D) For a process to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G < 0)$.
The relationship is given by $\Delta G = \Delta H - T\Delta S$.
For option $D$,$\Delta H$ is positive and $\Delta S$ is negative.
Therefore,$\Delta G = (+\Delta H) - T(-\Delta S) = (+\Delta H) + (T\Delta S)$.
Since both terms are positive,$\Delta G$ will always be positive at all temperatures.
Thus,the process will never be spontaneous.

(D) For a process to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,where $\Delta G = \Delta H - T\Delta S$.
If $\Delta H > 0$ (endothermic) and $\Delta S < 0$ (decrease in entropy),then $\Delta G = (\text{positive}) - T(\text{negative}) = \text{positive} + \text{positive} = \text{positive}$.
Since $\Delta G > 0$,the process is non-spontaneous (impossible) at all temperatures.

(C) The relationship for Gibbs energy is given by $\Delta G = \Delta H - T\Delta S$.
For $\Delta G = 0$,the equation becomes $\Delta H = T\Delta S$.
Given $\Delta H = 40.63 \ kJ/mol = 40630 \ J/mol$ and $\Delta S = 108.8 \ J/K \cdot mol$.
Substituting the values: $T = \frac{\Delta H}{\Delta S} = \frac{40630 \ J/mol}{108.8 \ J/K \cdot mol} = 373.4 \ K$.

(C) The chemical reaction is: $Fe(s) + 2HCl(aq) \to FeCl_2(aq) + H_2(g)$.
Number of moles of $Fe = \frac{50 \ g}{56 \ g/mol} = 0.8928 \ mol$.
Since $1 \ mol$ of $Fe$ produces $1 \ mol$ of $H_2$,moles of $H_2$ produced = $0.8928 \ mol$.
The volume of $H_2$ gas produced at $25 \ ^oC$ $(298 \ K)$ and $1 \ atm$ is calculated using the ideal gas law $PV = nRT$:
$V = \frac{nRT}{P} = \frac{0.8928 \ mol \times 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 298 \ K}{1 \ atm} = 21.842 \ L$.
Work done $(W)$ in an open beaker is given by $W = -P_{ext} \Delta V$.
Assuming the initial volume of the solid/liquid is negligible compared to the gas volume,$\Delta V \approx V_{gas} = 21.842 \ L$.
$W = -1 \ atm \times 21.842 \ L = -21.842 \ L \cdot atm$.
To convert $L \cdot atm$ to Joules $(J)$,use the conversion factor $1 \ L \cdot atm = 101.325 \ J$:
$W = -21.842 \times 101.325 \ J = -2213.13 \ J$.
Rounding to the nearest provided option,the correct answer is $-2212.39 \ J$.

(A) For a reaction to be in equilibrium,the change in Gibbs free energy $(\Delta G)$ must be zero.
Using the relation $\Delta G = \Delta H - T \Delta S$,at equilibrium $\Delta G = 0$,so $\Delta H = T \Delta S$.
Given $\Delta H = 30.5 \ kJ \ mol^{-1}$ and $\Delta S = 0.066 \ kJ \ K^{-1} \ mol^{-1}$.
Substituting the values: $30.5 = T \times 0.066$.
$T = \frac{30.5}{0.066} \approx 462.12 \ K$.

(B) According to the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$.
Given that $\Delta H < T \Delta S$,it follows that $\Delta H - T \Delta S < 0$.
Therefore,$\Delta G < 0$.

(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,i.e.,$\Delta G < 0$.
Given $\Delta G = \Delta H - T\Delta S$.
At equilibrium,$\Delta G = 0$,so $T = \frac{\Delta H}{\Delta S}$.
Substituting the values: $\Delta H = 179.1 \times 10^3 \, J \, mol^{-1}$ and $\Delta S = 160.2 \, J \, K^{-1} \, mol^{-1}$.
$T = \frac{179.1 \times 10^3}{160.2} \approx 1118 \, K$.
Since the reaction is endothermic $(\Delta H > 0)$ and entropy increases $(\Delta S > 0)$,the reaction becomes spontaneous at temperatures higher than the equilibrium temperature.
Therefore,the reaction becomes spontaneous above $1118 \, K$.

(B) The spontaneity of a reaction is determined by the Gibbs free energy change $(\Delta G)$,given by the equation: $\Delta G = \Delta H - T\Delta S$.
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: $\Delta G = (-11.7 \times 10^3) - (298 \times -105)$.
$\Delta G = -11700 + 31290 = +19590 \, J \, mol^{-1}$.
Since $\Delta G > 0$,the reaction is non-spontaneous.

(D) For a reaction to be spontaneous at all temperatures,the Gibbs free energy change $\Delta G = \Delta H - T\Delta S$ must be negative at all values of $T$.
This condition is satisfied when $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in entropy).
Looking at the given data:
Reaction $I$: $\Delta H > 0, \Delta S > 0$ (Spontaneous at high $T$)
Reaction $II$: $\Delta H < 0, \Delta S < 0$ (Spontaneous at low $T$)
Reaction $III$: $\Delta H > 0, \Delta S < 0$ (Non-spontaneous at all $T$)
Reaction $IV$: $\Delta H < 0, \Delta S > 0$ (Spontaneous at all $T$)
Reaction $V$: $\Delta H > 0, \Delta S = 0$ (Non-spontaneous at all $T$)
Therefore,Reaction $IV$ is spontaneous at all temperatures.

(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative.
$\Delta G = \Delta H - T\Delta S$.
Given that both $\Delta H$ and $\Delta S$ are positive $(+ve)$,
$\Delta G = (+ve) - T(+ve)$.
If $T\Delta S > \Delta H$,then the term $T\Delta S$ is larger than $\Delta H$,making $\Delta G$ negative.
Therefore,the reaction is spontaneous when $T\Delta S > \Delta H$.

(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G = \Delta H - T\Delta S$ must be negative $(\Delta G < 0)$.
Given that the reaction is endothermic,$\Delta H > 0$ (positive).
At the freezing point of water $(T = 273 \ K)$,the reaction is non-spontaneous,meaning $\Delta G > 0$,so $\Delta H > T\Delta S$.
At the boiling point of water $(T = 373 \ K)$,the reaction becomes feasible (spontaneous),meaning $\Delta G < 0$,so $\Delta H < T\Delta S$.
Since $\Delta H$ is positive,for $\Delta G$ to become negative as $T$ increases,$\Delta S$ must also be positive. Thus,both $\Delta H$ and $\Delta S$ are positive.

(B) The standard free energy change is calculated using the formula: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.
Given: $\Delta H^\circ = 58.04 \ kJ$,$\Delta S^\circ = 176.7 \ J/K = 0.1767 \ kJ/K$,and $T = 25 + 273 = 298 \ K$.
Substituting the values: $\Delta G^\circ = 58.04 \ kJ - (298 \ K \times 0.1767 \ kJ/K)$.
$\Delta G^\circ = 58.04 \ kJ - 52.66 \ kJ = 5.38 \ kJ$.

(C) The Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T\Delta S$.
At equilibrium,the free energy change $\Delta G = 0$.
Therefore,$\Delta H = T\Delta S$.
Substituting the given values: $30.56 \ kJ \ mol^{-1} = T \times 0.066 \ kJ \ K^{-1} \ mol^{-1}$.
Solving for $T$: $T = \frac{30.56}{0.066} \approx 463 \ K$.

(A) The condition for spontaneity is $\Delta G < 0$.
Given $\Delta G = \Delta H - T \Delta S$.
For the reaction to be spontaneous,$\Delta H - T \Delta S < 0$,which implies $T \Delta S > \Delta H$ or $T > \frac{\Delta H}{\Delta S}$.
Given $\Delta H = 35.5 \ kJ \ mol^{-1} = 35500 \ J \ mol^{-1}$ and $\Delta S = 83.6 \ J \ K^{-1} \ mol^{-1}$.
Calculating the equilibrium temperature: $T = \frac{35500}{83.6} \approx 424.64 \ K \approx 425 \ K$.
Since $\Delta H > 0$ and $\Delta S > 0$,the reaction is spontaneous at temperatures higher than the equilibrium temperature,i.e.,$T > 425 \ K$.

(D) The Gibbs free energy equation is given by $\Delta G = \Delta H - T \Delta S$.
For a reaction to be spontaneous at all temperatures,$\Delta G$ must be negative $(< 0)$ for all values of $T$.
If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in entropy),then $\Delta G = (\text{negative}) - T(\text{positive})$,which will always be negative regardless of the temperature $T$.
Therefore,the correct condition is $\Delta H < 0$ and $\Delta S > 0$.

(C) According to the Gibbs equation:
$\Delta G = \Delta H - T \Delta S$
When $\Delta G = 0$,the system is at equilibrium,so:
$\Delta H = T \Delta S$
Given:
$\Delta H = 40.63 \ kJ \ mol^{-1} = 40.63 \times 10^{3} \ J \ mol^{-1}$
$\Delta S = 108.8 \ J \ K^{-1} \ mol^{-1}$
Substituting the values:
$T = \frac{\Delta H}{\Delta S}$
$T = \frac{40.63 \times 10^{3} \ J \ mol^{-1}}{108.8 \ J \ K^{-1} \ mol^{-1}}$
$T \approx 373.43 \ K$
Thus,the temperature is approximately $373.4 \ K$.

(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G < 0)$.
Given: $\Delta H = 170 \ kJ = 170000 \ J$ and $\Delta S = 170 \ J \ K^{-1}$.
Using the relation $\Delta G = \Delta H - T \Delta S$,for spontaneity:
$\Delta H - T \Delta S < 0$
$\Delta H < T \Delta S$
$T > \frac{\Delta H}{\Delta S}$
$T > \frac{170000 \ J}{170 \ J \ K^{-1}}$
$T > 1000 \ K$.
Among the given options,the reaction becomes spontaneous at temperatures above $1000 \ K$,and $1110 \ K$ is the only value satisfying this condition.

(C) For a process occurring at constant temperature and pressure:
$1$. If $\Delta G_{system} < 0,$ the process is spontaneous.
$2$. If $\Delta G_{system} > 0,$ the process is non-spontaneous.
$3$. If $\Delta G_{system} = 0,$ the system is in a state of equilibrium.

(B) For the reaction $Br_{2(l)} + Cl_{2(g)} \rightarrow 2BrCl_{(g)}$,we have:
$\Delta H = 30 \ kJ \ mol^{-1} = 30000 \ J \ mol^{-1}$
$\Delta S = 105 \ J \ K^{-1} \ mol^{-1}$
At equilibrium,the Gibbs free energy change $\Delta G$ is $0$.
Using the relation $\Delta G = \Delta H - T \Delta S$,we set $\Delta G = 0$:
$0 = \Delta H - T \Delta S$
$\Delta H = T \Delta S$
$T = \frac{\Delta H}{\Delta S} = \frac{30000 \ J \ mol^{-1}}{105 \ J \ K^{-1} \ mol^{-1}} \approx 285.7 \ K$.

(B) At equilibrium,the Gibbs free energy change $\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T \Delta S$,at equilibrium we have $\Delta H - T_e \Delta S = 0$,which gives $T_e = \frac{\Delta H}{\Delta S}$.
For a reaction to be spontaneous,the condition is $\Delta G < 0$.
Substituting the expression for $\Delta G$,we get $\Delta H - T \Delta S < 0$,which implies $\Delta H < T \Delta S$.
Dividing by $\Delta S$ (since $\Delta S > 0$),we get $T > \frac{\Delta H}{\Delta S}$.
Since $T_e = \frac{\Delta H}{\Delta S}$,the condition for spontaneity is $T > T_e$.

(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 at all temperatures,$\Delta G$ must always be positive.
This occurs when $\Delta H$ is positive (endothermic) and $\Delta S$ is negative (decrease in entropy).
In option $D$,$\Delta H = +ve$ and $\Delta S = -ve$,so $\Delta G = (+ve) - T(-ve) = (+ve) + (T \times +ve)$,which is always positive.
Therefore,the reaction in option $D$ is never spontaneous.

(B) For an ideal gas,enthalpy $H = U + PV = nC_vT + nRT = n(C_v + R)T = nC_pT$.
Since $H = 10V^2$ and $H = nC_pT$,we have $nC_pT = 10V^2$.
For $He$ (monatomic),$C_p = \frac{5}{2}R$,so $2 \times \frac{5}{2}R \times T = 10V^2$,which simplifies to $5RT = 10V^2$,or $T = \frac{2}{R}V^2$.
This implies $T \propto V^2$,or $TV^{-2} = \text{constant}$.
Using the ideal gas law $PV = nRT$,we substitute $T = \frac{PV}{nR}$ into the relation: $\frac{PV}{nR}V^{-2} = \text{constant} \Rightarrow PV^{-1} = \text{constant} = K$.
Work done $w = -\int_{V_1}^{V_2} P \, dV = -\int_{V_1}^{V_2} K V \, dV = -\frac{K}{2}(V_2^2 - V_1^2)$.
Since $PV = nRT$,$K = PV^{-1} = \frac{nRT}{V^2}$.
Substituting $V^2 = \frac{RT}{2}$ (from $T = \frac{2}{R}V^2$),we get $K = \frac{nRT}{RT/2} = 2n = 2(2) = 4$.
Thus,$w = -\frac{4}{2}(V_2^2 - V_1^2) = -2(\frac{RT_2}{2} - \frac{RT_1}{2}) = -R(T_2 - T_1)$.
$w = -R(400 - 200) = -200 \, R$.

(B) For an isothermal reversible expansion,the work done is given by the formula: $w = -2.303 \times n \times R \times T \times log(\frac{V_2}{V_1})$
Given: $n = 1 \, \text{mole}$,$R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1}$,$T = 300 \, \text{K}$,$V_1 = 25 \, \text{L}$,$V_2 = 250 \, \text{L}$
Substituting the values: $w = -2.303 \times 1 \times 8.314 \times 300 \times log(\frac{250}{25})$
$w = -2.303 \times 8.314 \times 300 \times log(10)$
Since $\log(10) = 1$,$w = -2.303 \times 8.314 \times 300 \approx -5744 \, \text{J}$

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G < 0)$.
Given $\Delta G = \Delta H - T\Delta S < 0$.
Substituting the values: $75000 \ J \ mol^{-1} - T(120 \ J \ K^{-1} \ mol^{-1}) < 0$.
$75000 < 120T$.
$T > \frac{75000}{120}$.
$T > 625 \ K$.
Thus,the reaction is spontaneous at temperatures greater than $625 \ K$.

(B) The reaction $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ represents the phase transition of water at its boiling point.
At $373 \ K$ and $1 \ atm$,the system is at equilibrium.
The Gibbs free energy change is given by the equation $\Delta G = \Delta H - T \Delta S$.
Since the system is at equilibrium,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 $\Delta H = T \Delta S$.

(A) The reaction is: $U_{(s)} + 2Br_2(\ell) \to UBr_{4(s)}$.
First,calculate the change in entropy for the reaction,$\Delta S^o_{rxn}$:
$\Delta S^o_{rxn} = \sum S^o_{products} - \sum S^o_{reactants}$
$\Delta S^o_{rxn} = S^o(UBr_{4(s)}) - [S^o(U_{(s)}) + 2 \times S^o(Br_2(\ell))]$
$\Delta S^o_{rxn} = 242.6 - [50.3 + 2 \times 152.3] \ J/K \cdot mol$
$\Delta S^o_{rxn} = 242.6 - [50.3 + 304.6] = 242.6 - 354.9 = -112.3 \ J/K \cdot mol = -0.1123 \ kJ/K \cdot mol$.
Using the Gibbs-Helmholtz equation: $\Delta G^o = \Delta H^o - T\Delta S^o$.
Assuming standard temperature $T = 298 \ K$:
$-788.6 \ kJ = \Delta H^o - (298 \ K \times -0.1123 \ kJ/K \cdot mol)$
$-788.6 = \Delta H^o + 33.4654$
$\Delta H^o = -788.6 - 33.4654 = -822.0654 \ kJ/mol \approx -822.1 \ kJ/mol$.

(B) The spontaneity of a process is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For a process to be spontaneous at all temperatures,$\Delta G$ must be negative $(\Delta G < 0)$.
If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in entropy),then $\Delta G = (\text{negative}) - T(\text{positive})$.
Since $T$ is always positive in Kelvin,$\Delta G$ will always be negative regardless of the temperature $T$.

(A) 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,$\Delta G$ must be negative $(\Delta G < 0)$.
If $\Delta H > 0$ (endothermic) and $\Delta S > 0$ (increase in entropy),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.
At low temperatures,$T\Delta S < \Delta H$,so $\Delta G$ is positive,making the reaction non-spontaneous.

(C) For a process to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,i.e.,$\Delta G < 0$.
We know that $\Delta G = \Delta H - T\Delta S$.
Setting $\Delta G = 0$ gives the equilibrium temperature: $0 = \Delta H - T\Delta S$.
Therefore,$T = \frac{\Delta H}{\Delta S}$.
Substituting the given values: $T = \frac{200 \ J \ mol^{-1}}{40 \ J \ K^{-1} \ mol^{-1}} = 5 \ K$.
For the process to be spontaneous,$T$ must be greater than $5 \ K$ (since $\Delta H$ and $\Delta S$ are both positive,the process is spontaneous at higher temperatures).

(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G^o$ must be negative.
The relationship is given by $\Delta G^o = \Delta H^o - T \Delta S^o$.
At equilibrium,$\Delta G^o = 0$,so $T = \frac{\Delta H^o}{\Delta S^o}$.
Given $\Delta H^o = 491.1 \ kJ \ mol^{-1} = 491100 \ J \ mol^{-1}$ and $\Delta S^o = 198.0 \ J \ K^{-1} \ mol^{-1}$.
$T = \frac{491100 \ J \ mol^{-1}}{198.0 \ J \ K^{-1} \ mol^{-1}} = 2480.3 \ K$.
Therefore,the reaction becomes spontaneous at temperatures above $2480.3 \ K$.

(A) The standard Gibbs energy equation is given by $\Delta G^{o} = \Delta H^{o} - T\Delta S^{o}$.
Comparing this with the given equation $\Delta G^{o} = A - BT$,we find that $A = \Delta H^{o}$ and $B = \Delta S^{o}$.
For an endothermic reaction,the enthalpy change $\Delta H^{o}$ is positive,which implies $A > 0$.
For an exothermic reaction,the enthalpy change $\Delta H^{o}$ is negative,which implies $A < 0$.
Therefore,the reaction is endothermic if $A > 0$.

(B) For a process to be spontaneous,the change in Gibbs free energy must be negative,i.e.,$\Delta G < 0$.
We use the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T\Delta S$.
For the process to be spontaneous at all temperatures $(T)$,the term $\Delta G$ must remain negative regardless of the value of $T$.
If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in entropy),then $\Delta G = (\text{negative}) - T(\text{positive})$.
Since $T$ is always positive in Kelvin,$\Delta G$ will always be negative at all temperatures.

(A) For a spontaneous reaction,the Gibbs free energy change $ \Delta G $ must be negative,where $ \Delta G = \Delta H - T \Delta S $.
If $ \Delta H $ is positive and $ \Delta S $ is negative,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 cannot occur spontaneously under any conditions.

(B) The reaction is: $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$.
For $1 \, mol$ of $Zn$,the change in gaseous moles is $\Delta n_g = 1 \, mol \, (H_2) - 0 = 1$.
For $2 \, mol$ of $Zn$,$\Delta n_g = 2 \, mol$.
The work done in an open vessel is given by $W = -\Delta n_g RT$.
Given $T = 25 \, ^oC = 298 \, K$ and $R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
$W = -2 \times 8.314 \times 298 = -4955.144 \, J$.
Converting to $kJ$,$W = -4.955 \, kJ$.

(B) Solution:- $(B)$ $\Delta G > 0$
As we know that the melting of ice at a temperature below $273 \ K$ (i.e.,$-15\,^{\circ}C$) is a non-spontaneous process.
For a non-spontaneous process,the change in Gibbs free energy $(\Delta G)$ is always greater than $0$ (i.e.,$\Delta G > 0$).

(B) The melting of ice is a phase transition process represented as $H_2O(s) \rightleftharpoons H_2O(l)$.
For any process,the Gibbs free energy change is given by $\Delta G = \Delta H - T\Delta S$.
At temperatures below the melting point ($0\,^{\circ}C$ or $273.15\,K$),the process of melting is non-spontaneous.
At $-15\,^{\circ}C$ $(258.15\,K)$,the ice is stable and does not melt spontaneously.
Since the process is non-spontaneous at this temperature,$\Delta G$ must be positive $(\Delta G > 0)$.

(A) The balanced chemical equation is: $2H_2O_{2(l)} \to O_{2(g)} + 2H_2O_{(l)}$.
Molar mass of $H_2O_2 = 34 \ g \cdot mol^{-1}$.
Moles of $H_2O_2 = \frac{4.5 \ g}{34 \ g \cdot mol^{-1}} \approx 0.13235 \ mol$.
From the stoichiometry,$2 \ mol$ of $H_2O_2$ produces $1 \ mol$ of $O_{2(g)}$.
Therefore,moles of $O_{2(g)}$ produced $(\Delta n_g) = \frac{0.13235}{2} = 0.066175 \ mol$.
Work done $w = -P_{ext} \Delta V = -\Delta n_g RT$.
Using $R = 8.314 \ J \cdot K^{-1} \cdot mol^{-1}$ and $T = 298 \ K$:
$w = -(0.066175 \ mol) \times (8.314 \ J \cdot K^{-1} \cdot mol^{-1}) \times (298 \ K) \approx -163.9 \ J$.
Rounding to two significant figures,$w = -1.64 \times 10^2 \ J$.

(B) The formula for maximum work done in an isothermal reversible expansion is $W_{max} = -2.303 \ nRT \log(\frac{V_2}{V_1})$.
Given: Mass of $O_2 = 16 \ g$,Molar mass of $O_2 = 32 \ g/mol$.
Number of moles $n = \frac{16}{32} = 0.5 \ mol$.
Temperature $T = 300 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Initial volume $V_1 = 5 \ dm^3$,Final volume $V_2 = 25 \ dm^3$.
Substituting the values: $W_{max} = -2.303 \times 0.5 \times 8.314 \times 300 \times \log(\frac{25}{5})$.
$W_{max} = -2.303 \times 0.5 \times 8.314 \times 300 \times \log(5)$.
$W_{max} = -2.303 \times 0.5 \times 8.314 \times 300 \times 0.699$.
$W_{max} \approx -2009.6 \ J \approx -2010 \ J$.

(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,$\Delta G$ must be negative $(\Delta G < 0)$.
If $\Delta H$ is negative (exothermic) and $\Delta S$ is positive (increase in entropy),then $\Delta G = (\text{negative}) - T(\text{positive})$.
Since $T$ (absolute temperature) is always positive,$-T\Delta S$ will be negative.
Therefore,$\Delta G$ will always be negative regardless of the temperature,making the reaction always spontaneous.

(A) The formula for maximum work done in a reversible isothermal expansion is $w_{max} = -2.303 \ nRT \ \log_{10} \frac{V_2}{V_1}$.
Given: Mass of $O_2 = 16 \ g$,Molar mass of $O_2 = 32 \ g/mol$.
Number of moles $n = \frac{16}{32} = 0.5 \ mol$.
Temperature $T = 300 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$V_1 = 5 \ dm^3$,$V_2 = 25 \ dm^3$.
$w_{max} = -2.303 \times 0.5 \times 8.314 \times 300 \times \log_{10} \frac{25}{5}$.
$w_{max} = -2.303 \times 0.5 \times 8.314 \times 300 \times 0.699$.
$w_{max} \approx -2008.5 \ J \approx -2.01 \times 10^3 \ J$.
Since the work is done by the system,the magnitude of work done is $2.01 \times 10^3 \ J$.

(B) For a reaction to occur reversibly,the change in Gibbs free energy must be zero,i.e.,$\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T\Delta S$,we set $\Delta H - T\Delta S = 0$.
Given $\Delta H_r = 100\, kJ/mol = 100,000\, J/mol$ and $\Delta S_r = 400\, J/mol\cdot K$.
Substituting the values: $100,000 - T(400) = 0$.
$T = \frac{100,000}{400} = 250\, K$.

(B) The chemical reaction is: $Zn_{(s)} + 2HCl_{(aq)} \to ZnCl_{2(aq)} + H_{2(g)}$
For $2 \, mol$ of $Zn$,the number of moles of gas produced is $\Delta n_g = 1 \times 2 = 2 \, mol$.
The work done in an open vessel is given by $w = -P \Delta V = -\Delta n_g RT$.
Given $T = 25 + 273 = 298 \, K$ and $R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
$w = -2 \times 8.314 \times 298 \, J = -4959.15 \, J \approx -4.955 \, kJ$.

(C) The spontaneity of a reaction is determined by the Gibbs free energy change,$\Delta G = \Delta H - T\Delta S$.
At equilibrium,$\Delta G = 0$,so $T = \frac{\Delta H}{\Delta S}$.
Given $\Delta H = -40,000 \, J$ and $\Delta S = -50 \, J/K$,the equilibrium temperature is $T = \frac{-40,000 \, J}{-50 \, J/K} = 800 \, K$.
For $\Delta H < 0$ and $\Delta S < 0$,the reaction is spontaneous when $T < 800 \, K$ and non-spontaneous when $T > 800 \, K$.
Thus,the transition occurs around $800 \, K$.

(A) At equilibrium,the change in Gibbs free energy $\Delta G = 0$.
Using the relation $\Delta G = \Delta H - T\Delta S$,we get $\Delta H = T\Delta S$.
Therefore,$T = \frac{\Delta H_{vap}}{\Delta S_{vap}}$.
Given $\Delta H_{vap} = 40.73 \, kJ \, mol^{-1} = 40730 \, J \, mol^{-1}$ and $\Delta S_{vap} = 109 \, J \, K^{-1} \, mol^{-1}$.
$T = \frac{40730 \, J \, mol^{-1}}{109 \, J \, K^{-1} \, mol^{-1}} \approx 373.67 \, K$.
Converting to Celsius: $T(^circ C) = 373.67 - 273 = 100.67 \, ^\circ C$.

(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G < 0)$.
Given $\Delta G = \Delta H - T \cdot \Delta S$.
At equilibrium,$\Delta G = 0$,so $T = \frac{\Delta H}{\Delta S}$.
Converting $\Delta H$ to Joules: $\Delta H = 178.3 \times 1000 \, J = 178300 \, J$.
$T = \frac{178300 \, J}{160 \, J \cdot K^{-1}} = 1114.375 \, K \approx 1114 \, K$.
Since the reaction is endothermic $(\Delta H > 0)$ and entropy increases $(\Delta S > 0)$,the reaction becomes spontaneous at temperatures higher than the equilibrium temperature.
Therefore,the reaction becomes spontaneous at $T > 1114 \, K$.