Question:
Consider the following reaction: $ \text{CO}(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}_2(g) $ At 27°C, the standard entropy change of the process becomes -0.094 kJ/mol·K. Moreover, standard free energies for the formation of $ \text{CO}_2(g) $ and $ \text{CO}(g) $ are -394.4 and -137.2 kJ/mol, respectively. Predict the nature of the above chemical reaction.
Consider the following reaction: $ \text{CO}(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}_2(g) $ At 27°C, the standard entropy change of the process becomes -0.094 kJ/mol·K. Moreover, standard free energies for the formation of $ \text{CO}_2(g) $ and $ \text{CO}(g) $ are -394.4 and -137.2 kJ/mol, respectively. Predict the nature of the above chemical reaction.
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For a reaction to be spontaneous, the Gibbs free energy change must be negative. This is typically true for exothermic reactions with a negative entropy change.
Updated On: Jan 5, 2026
- Exothermic and spontaneous
- Endothermic and spontaneous
- Exothermic and non-spontaneous (driven)
- Exothermic and equilibrating
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The Correct Option is A
Solution and Explanation
The reaction involves the combustion of carbon monoxide (\( \text{CO} \)) to form carbon dioxide (\( \text{CO}_2 \)).
The standard entropy change (\( \Delta S \)) is negative, indicating a decrease in disorder, suggesting a release of energy.
The standard free energy change (\( \Delta G \)) for the reaction can be calculated using the formula:
\[
\Delta G = \Delta H - T\Delta S
\]
Since the reaction is exothermic, it is spontaneous at 27°C, as the decrease in Gibbs free energy (\( \Delta G \)) drives the reaction forward.
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