Question

For the reaction at 300 K, ; The value of is:

• A. -32.6 kJ/mol
• B. -59.6 kJ/mol
• C. -92.2 kJ/mol
• D. 27.0 kJ/mol

Answer 1

Answer: (A)

-32.6 kJ/mol

Answer 2

The Gibbs free energy change ($\Delta G^\circ$) is calculated using the equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ Given: $T = 300$ K $\Delta H^\circ = -92.2$ kJ/mol $\Delta S^\circ = -198.7$ J/mol·K

First, convert $\Delta S^\circ$ to kJ/mol·K: $\Delta S^\circ = -198.7 \text{ J/mol·K} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = -0.1987 \text{ kJ/mol·K}$

Now, substitute the values into the Gibbs free energy equation: $\Delta G^\circ = (-92.2 \text{ kJ/mol}) - (300 \text{ K}) \times (-0.1987 \text{ kJ/mol·K})$ $\Delta G^\circ = -92.2 \text{ kJ/mol} - (-59.61 \text{ kJ/mol})$ $\Delta G^\circ = -92.2 \text{ kJ/mol} + 59.61 \text{ kJ/mol}$ $\Delta G^\circ = -32.59 \text{ kJ/mol}$

Considering significant figures, $\Delta H^\circ$ has one decimal place (-92.2) and $T\Delta S^\circ$ calculated as $300 \times -0.1987 = -59.61$. When multiplying, we consider the least number of significant figures. Assuming 300 K has 3 significant figures, the result of $T\Delta S^\circ$ should also be rounded to 3 significant figures, which is -59.6 kJ/mol.

Performing the subtraction with appropriate decimal places: $\Delta G^\circ = -92.2 \text{ kJ/mol} - (-59.6 \text{ kJ/mol})$ $\Delta G^\circ = -92.2 \text{ kJ/mol} + 59.6 \text{ kJ/mol}$ $\Delta G^\circ = -32.6 \text{ kJ/mol}$