Let us consider a reversible reaction at temperature, T . In this reaction, both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ were observed to have positive values. If the equilibrium temperature is $\mathrm{T}_{\mathrm{e}}$, then the reaction becomes spontaneous at:
Let us consider a reversible reaction at temperature, T . In this reaction, both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ were observed to have positive values. If the equilibrium temperature is $\mathrm{T}_{\mathrm{e}}$, then the reaction becomes spontaneous at:
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- $\mathrm{T}=\mathrm{T}_{\mathrm{e}}$
- $\mathrm{T}_{\mathrm{e}}>\mathrm{T}$
- $\mathrm{T}>\mathrm{T}_{\mathrm{e}}$
- $\mathrm{T}_{\mathrm{e}}=5 \mathrm{~T}$
The Correct Option is C
Approach Solution - 1
To determine the conditions under which a reaction becomes spontaneous, we must apply the concept of Gibbs free energy change, denoted as \(\Delta G\). The equation for Gibbs free energy is:
\(\Delta G = \Delta H - T \Delta S\),
where:
- \(\Delta G\) is the change in Gibbs free energy
- \(\Delta H\) is the change in enthalpy
- \(\Delta S\) is the change in entropy
- \(T\) is the temperature in Kelvin
For a reaction to be spontaneous, \(\Delta G\) must be negative. Given that both \(\Delta H\) and \(\Delta S\) are positive, the reaction can become spontaneous at higher temperatures.
To find the equilibrium temperature, \(\Delta G\) is set to zero:
\(\Delta H = T_e \Delta S\)
At this temperature, \(\Delta G = 0\) and the reaction is at equilibrium. For the reaction to be spontaneous, the condition is:
\(\Delta G = \Delta H - T \Delta S < 0\)
Rewriting the inequality:
\(T \Delta S > \Delta H\), implying that \(T > T_e\)
This means that the reaction becomes spontaneous when the temperature is greater than the equilibrium temperature, \(T_e\).
Thus, the correct answer is:
\(T > T_e\)
Approach Solution -2
2. Given that both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are positive: \[ \Delta \mathrm{G} = \Delta \mathrm{H} - \mathrm{T} \Delta \mathrm{S}<0 \] \[ \mathrm{T}>\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}} = \mathrm{T}_{\mathrm{e}} \] Therefore, the correct answer is (3) $\mathrm{T}>\mathrm{T}_{\mathrm{e}}$.
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