Question

For a given reaction, $\Delta H =35.5\, kJ\, mol ^{-1}$ and $\Delta S =83.6\, JK ^{-1}\, mol ^{-1} .$ The reaction is spontaneous at $:$ (Assume that $\Delta H$ and $\Delta S$ do not vary with temperature)

• A. T > 425 K
• B. All temperatures
• C. T > 298 K
• D. T < 425 K

Answer 1

The Correct Option is A

To determine the temperatures at which the given reaction is spontaneous, we can use the Gibbs free energy equation:

\Delta G = \Delta H - T \Delta S

A reaction is spontaneous when the Gibbs free energy change, \Delta G, is negative: \Delta G < 0. This translates to the following inequality for the given reaction:

\Delta H - T \Delta S < 0

Plugging in the given values, where \Delta H = 35.5 \, \text{kJ mol}^{-1} = 35500 \, \text{J mol}^{-1} and \Delta S = 83.6 \, \text{J K}^{-1} \text{mol}^{-1}, we get:

35500 \, \text{J mol}^{-1} - T \cdot 83.6 \, \text{J K}^{-1} \text{mol}^{-1} < 0

Simplifying this, we find:

35500 < T \cdot 83.6

T > \frac{35500}{83.6}

Calculating the above division:

T > 425 \, \text{K}

Hence, the reaction is spontaneous at temperatures greater than 425 K.

To conclude, the correct answer is: T > 425 K. This excludes all other options as they do not satisfy the condition for spontaneity based on the values provided.