The entropy and enthalpy changes for the reaction \( \text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{H}_2\text{(g)} \) at 300 K and 1 atm are respectively \(-42.4 \text{ JK}^{-1}\) and \(-41.2 \text{ kJ}\). The temperature at which the reaction will go in the reverse direction is
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The sign of \(\Delta G\) determines the direction of a reaction. The equilibrium temperature \(T_{eq} = \Delta H / \Delta S\) is the "tipping point". If \(\Delta H\) and \(\Delta S\) have the same sign, the spontaneity depends on temperature.
- If both are negative (as here), the reaction is spontaneous at low T (\(<T_{eq}\)) and non-spontaneous at high T (\(>T_{eq}\)).
- If both are positive, the reverse is true.
The spontaneity of a reaction is determined by the sign of the Gibbs free energy change, \( \Delta G = \Delta H - T\Delta S \).
The reaction goes in the forward direction if \( \Delta G<0 \).
The reaction goes in the reverse direction if \( \Delta G>0 \).
The reaction is at equilibrium if \( \Delta G = 0 \).
Let's find the temperature (T_eq) at which the reaction is at equilibrium.
At equilibrium, \( \Delta G = 0 \), so \( \Delta H - T_{eq}\Delta S = 0 \).
\( T_{eq} = \frac{\Delta H}{\Delta S} \).
We are given:
Enthalpy change, \( \Delta H = -41.2 \text{ kJ} = -41200 \text{ J} \).
Entropy change, \( \Delta S = -42.4 \text{ JK}^{-1} \).
It is important to use consistent units (Joules).
\( T_{eq} = \frac{-41200}{-42.4} \approx 971.7 \) K.
Now, let's analyze the spontaneity based on temperature.
\( \Delta G = -41200 - T(-42.4) = -41200 + 42.4T \).
The reaction will go in the reverse direction when \( \Delta G>0 \).
\( -41200 + 42.4T>0 \).
\( 42.4T>41200 \).
\( T>\frac{41200}{42.4} \).
\( T>971.7 \) K.
The reaction will be spontaneous in the reverse direction at temperatures above the equilibrium temperature.
Looking at the options, the only temperature greater than 971.7 K is 971.8 K.
