For the reaction at $298 \, K$,$2 A + B \rightarrow C$. Given $\Delta H = 400 \, kJ \, mol^{-1}$ and $\Delta S = 0.2 \, kJ \, K^{-1} \, mol^{-1}$. At what temperature will the reaction become spontaneous,considering $\Delta H$ and $\Delta S$ to be constant over the temperature range?
(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 $T > 2000 \, K$.
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 $T > 2000 \, K$.
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Similar Questions
For the reaction $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$,$\Delta H = -285.8 \ kJ \ mol^{-1}$,$\Delta S = -0.163 \ kJ \ mol^{-1} K^{-1}$. What is the value of free energy change at $27 \ ^\circ C$ for the reaction in $kJ \ mol^{-1}$?
Medium
View SolutionData given for the following reaction is as follows:
$FeO_{(s)} + C_{(\text{graphite})} \longrightarrow Fe_{(s)} + CO_{(g)}$
Substance
$\Delta H^{\circ} \text{ (kJ mol}^{-1})$
$\Delta S^{\circ} \text{ (J mol}^{-1} \text{ K}^{-1})$
$FeO_{(s)}$
$-266.3$
$57.49$
$C_{(\text{graphite})}$
$0$
$5.74$
$Fe_{(s)}$
$0$
$27.28$
$CO_{(g)}$
$-110.5$
$197.6$
The minimum temperature in $K$ at which the reaction becomes spontaneous is ....... .
(Integer answer)
$FeO_{(s)} + C_{(\text{graphite})} \longrightarrow Fe_{(s)} + CO_{(g)}$
| Substance | $\Delta H^{\circ} \text{ (kJ mol}^{-1})$ | $\Delta S^{\circ} \text{ (J mol}^{-1} \text{ K}^{-1})$ |
| $FeO_{(s)}$ | $-266.3$ | $57.49$ |
| $C_{(\text{graphite})}$ | $0$ | $5.74$ |
| $Fe_{(s)}$ | $0$ | $27.28$ |
| $CO_{(g)}$ | $-110.5$ | $197.6$ |
The minimum temperature in $K$ at which the reaction becomes spontaneous is ....... .
(Integer answer)
The standard entropy change for the reaction $4 Fe ( s ) + 3 O_{2} ( g ) \rightarrow 2 Fe_{2} O_{3} ( s )$ is $-550 \ J K^{-1}$ at $298 \ K$. Given: The standard enthalpy change for the reaction is $-165 \ kJ \ mol^{-1}$. The temperature in $K$ at which the reaction attains equilibrium is ... . (Nearest Integer)
The maximum work done in expanding $16 \ g$ of oxygen at $300 \ K$ from an initial volume of $5 \ dm^3$ to a final volume of $25 \ dm^3$ isothermally and reversibly is $(\log_{10} 5 = 0.699)$.
Difficult
View SolutionThe correct thermodynamic conditions for a spontaneous reaction at all temperatures are:
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