Question

calculate \\( \delta g^{\circ} \\) for the reaction

\\( \mathrm{n}_{2} \mathrm{o}_{4}(\mathrm{~g}) \
ightleftharpoons 2 \mathrm{no}_{2}(\mathrm{~g}) \\)

\\( \mathrm{k}=0.144 \\) at \\( 298 \mathrm{~k} \\).

Explanation:

Step1: Recall the formula

The formula relating $\Delta G^{\circ}$ and $K$ is $\Delta G^{\circ}=-RT\ln K$, where $R = 8.314\text{ J/(mol}\cdot\text{K)}$ and $T$ is the temperature in Kelvin.

Step2: Substitute the values

Substitute $R = 8.314\text{ J/(mol}\cdot\text{K)}$, $T = 298\text{ K}$, and $K=0.144$ into the formula.
First, calculate $\ln K=\ln(0.144)\approx - 1.938$.
Then, $\Delta G^{\circ}=-(8.314\text{ J/(mol}\cdot\text{K)})\times(298\text{ K})\times(- 1.938)$.

Step3: Calculate the result

$\Delta G^{\circ}=(8.314\times298\times1.938)\text{ J/mol}\approx4.80\times10^{3}\text{ J/mol}$.

Answer:

$$\Delta G^{\circ}=4.80\times10^{3}\text{ J/mol}$$