Question

determine $\delta g^\circ$ for the reaction at $25\\ ^\circ\text{c}$.
$\ce{2 no2 (g) -> 2 no (g) + o2 (g)}$

substance	$\delta g^\circ f$ (kj/mol)
$\ce{no (g)}$	86.6

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 $\Delta G^\circ=\sum n\Delta G^\circ_f(\text{products})-\sum m\Delta G^\circ_f(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients of products and reactants respectively.

Step2: Identify products and reactants with their coefficients

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

• Reactants: $\text{NO}_2(\text{g})$ with coefficient $2$, $\Delta G^\circ_f(\text{NO}_2(\text{g})) = 51.3\ \text{kJ/mol}$
• Products: $\text{NO}(\text{g})$ with coefficient $2$, $\Delta G^\circ_f(\text{NO}(\text{g})) = 86.6\ \text{kJ/mol}$; $\text{O}_2(\text{g})$ with coefficient $1$, and for elements in their standard state, $\Delta G^\circ_f(\text{O}_2(\text{g})) = 0\ \text{kJ/mol}$ (since the standard Gibbs free energy of formation of an element in its standard state is zero).

Step3: Calculate the sum of $\Delta G^\circ_f$ for products

$\sum n\Delta G^\circ_f(\text{products})=2\times\Delta G^\circ_f(\text{NO}(\text{g})) + 1\times\Delta G^\circ_f(\text{O}_2(\text{g}))$
Substitute the values: $2\times86.6 + 1\times0=173.2\ \text{kJ}$

Step4: Calculate the sum of $\Delta G^\circ_f$ for reactants

$\sum m\Delta G^\circ_f(\text{reactants})=2\times\Delta G^\circ_f(\text{NO}_2(\text{g}))$
Substitute the value: $2\times51.3 = 102.6\ \text{kJ}$

Step5: Calculate $\Delta G^\circ$

$\Delta G^\circ=\sum n\Delta G^\circ_f(\text{products})-\sum m\Delta G^\circ_f(\text{reactants})$
Substitute the values from Step3 and Step4: $\Delta G^\circ=173.2 - 102.6=70.6\ \text{kJ}$

Answer:

The value of $\Delta G^\circ$ for the reaction at $25^\circ\text{C}$ is $\boldsymbol{70.6\ \text{kJ}}$ (per mole of reaction as written).