Question

given $\delta h^\circ = -19.4\\,\text{kj/mol}$ and $\delta s^\circ = +92.3\\,\text{j} \cdot \text{mol}^{-1}\text{k}^{-1}$, determine $\delta g^\circ$, in kj/mol, for the following phase change at $25.0\\,^\circ\text{c}$.\
\\\text{co}_2(\text{g}) \
ightarrow \text{co}_2(\text{aq})\\

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 \(\Delta S^\circ = +92.3 \text{ J·mol}^{-1}\text{K}^{-1}\), we convert it to kJ by dividing by 1000: \(\Delta S^\circ = \frac{92.3}{1000} = 0.0923 \text{ kJ·mol}^{-1}\text{K}^{-1}\).

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: \(\Delta H^\circ = -19.4 \text{ kJ/mol}\), \(T = 298.15 \text{ K}\), and \(\Delta S^\circ = 0.0923 \text{ kJ·mol}^{-1}\text{K}^{-1}\).
\[

$$\begin{align*} \Delta G^\circ &= -19.4 - (298.15 \times 0.0923)\\ &= -19.4 - 27.52\\ &= -46.92 \end{align*}$$

\]

Answer:

\(\Delta G^\circ \approx -46.9 \text{ kJ/mol}\) (rounded to three significant figures)