Question

calculate $\delta g^\circ$ for the reaction below at $25\\ ^\circ\text{c}$.
$2\\ \text{no}_2\\ (\text{g})\\ \
ightarrow\\ \text{n}_2\text{o}_4\\ (\text{g})$

substance	$\delta g^\circ f\\ (\text{kj/mol})$
$\text{n}_2\text{o}_4\\ (\text{g})$	$97.8$

Explanation:

Step1: Recall the formula for $\Delta G^\circ$

The formula for the standard Gibbs free energy change of a reaction ($\Delta G^\circ$) is $\Delta G^\circ=\sum n\Delta G_f^\circ(\text{products})-\sum m\Delta G_f^\circ(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients of products and reactants respectively.

Step2: Identify the stoichiometric coefficients and $\Delta G_f^\circ$ values

For the reaction $2\text{NO}_2(\text{g})
ightarrow\text{N}_2\text{O}_4(\text{g})$:

• Reactant: $\text{NO}_2(\text{g})$ with stoichiometric coefficient $2$ and $\Delta G_f^\circ = 51.3\ \text{kJ/mol}$.
• Product: $\text{N}_2\text{O}_4(\text{g})$ with stoichiometric coefficient $1$ and $\Delta G_f^\circ = 97.8\ \text{kJ/mol}$.

Step3: Calculate $\sum n\Delta G_f^\circ(\text{products})$

For the product $\text{N}_2\text{O}_4(\text{g})$, $n = 1$, so $\sum n\Delta G_f^\circ(\text{products})=1\times97.8\ \text{kJ/mol}=97.8\ \text{kJ/mol}$.

Step4: Calculate $\sum m\Delta G_f^\circ(\text{reactants})$

For the reactant $\text{NO}_2(\text{g})$, $m = 2$, so $\sum m\Delta G_f^\circ(\text{reactants})=2\times51.3\ \text{kJ/mol}=102.6\ \text{kJ/mol}$.

Step5: Calculate $\Delta G^\circ$

Using the formula $\Delta G^\circ=\sum n\Delta G_f^\circ(\text{products})-\sum m\Delta G_f^\circ(\text{reactants})$, we substitute the values:
$\Delta G^\circ=97.8\ \text{kJ/mol}- 102.6\ \text{kJ/mol}=- 4.8\ \text{kJ/mol}$.

Answer:

$\boldsymbol{-4.8\ \text{kJ/mol}}$