Question

given $\delta h^\circ = +90.7\\ \text{kj/mol}$ and $\delta s^\circ = +221.0\\ \text{j} \cdot \text{mol}^{-1}\text{k}^{-1}$, determine $\delta g^\circ$, in kj/mol, for the following reaction at $25.0\\ ^\circ\text{c}$.\
\
$\text{ch}_3\text{oh}(\text{g}) \
ightarrow \text{co}_2(\text{g}) + 2\\ \text{h}_2(\text{g})$

Explanation:

Step1: Convert temperature to Kelvin

The temperature is given as \(25.0^\circ \text{C}\). To convert to Kelvin, we use the formula \(T = 25.0 + 273.15 = 298.15 \text{ K}\).

Step2: Convert \(\Delta S^\circ\) to kJ units

Given \(\Delta S^\circ = +221.0 \text{ J·mol}^{-1}\text{K}^{-1}\), we convert to kJ by dividing by 1000: \(\Delta S^\circ = \frac{221.0}{1000} = 0.2210 \text{ kJ·mol}^{-1}\text{K}^{-1}\).

Step3: Use the Gibbs free energy formula

The formula for Gibbs free energy is \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\). Substituting the values: \(\Delta H^\circ = 90.7 \text{ kJ/mol}\), \(T = 298.15 \text{ K}\), and \(\Delta S^\circ = 0.2210 \text{ kJ·mol}^{-1}\text{K}^{-1}\).
\[

$$\begin{align*} \Delta G^\circ &= 90.7 - (298.15 \times 0.2210)\\ &= 90.7 - 65.9\\ &= 24.8 \text{ kJ/mol} \end{align*}$$

\]

Answer:

\(24.8 \text{ kJ/mol}\)