Question

Select the correct answer,
What is the AG for the following reaction at 25°C?
N204(8) 2N02(9)
Given:
N20409): AG= 98.28 kJ/mol
NO2(9): AG= 51.8 kJ/mol
OA 5.32 kJ/mol
OB. 10.64 kJ/mol
OC -5.32 kj/mol
OD. -10.64 kJ/mol

Answer 1

Answer:

Choice A: $5.32\; \rm kJ \cdot mol^{-1}$.

Explanation:

The Gibbs Energy change in a reaction $\Delta G_\text{rxn}$ can be found from the Gibbs Energy of formation $\Delta G_f$:

$\begin{aligned}& \Delta G_\text{rxn}^{\circ} \\ &= \text{sum of \Delta G^\circ_f for products} \\ &\phantom{=} \; - \text{sum of \Delta G^\circ_f for reactants} \end{aligned}$.

In this case,

$\begin{aligned}& \Delta G_\text{rxn} \\ &= \text{sum of \Delta G for products} \\ &\phantom{=} \; - \text{sum of \Delta G_f for reactants} \\ &= 2 \,\Delta G_f\left(\mathrm{NO_2\, (g)}\right) - \Delta G_f(\mathrm{N_2O_4}) \\ &= 2 \times 51.8 - 98.28 \\ &= 5.32\; \rm kJ \cdot mol^{-1}\end{aligned}$.

Note that since there are two $\rm NO_2\, (g)$ molecules on the product side of this reaction, the $\Delta G_f$ for

One way to interpret this equation is to think about what $\Delta G_f$ means.

Refer to the diagram attached. The $\Delta G_f$ value of a substance is the energy it takes to form one mole of it from its elements under the standard conditions.

In this case, the two $\Delta G_f$ are like vectors that points from the elements to the two substances. However, the problem is asking for $\Delta G_\text{rxn}$, which is the dashed arrow that points from $\mathrm{N_2O_4\, (g)}$ to $\mathrm{2\, NO_2\, (g)}$. To find that value, invert $\Delta G_f(\mathrm{N_2O_4\, (g)})$ (the arrow on the left-hand side) and add that to $2\, \Delta G_f(\mathrm{NO_2\, (g)})$ (the arrow on the right-hand side.) The resultant value would be the dashed arrow, which stands for $\Delta G_\text{rxn}$. In other words, $\Delta G_\text{rxn} = (- \Delta G_f(\mathrm{N_2O_4\, (g)})) + 2\, \Delta G_f(\mathrm{NO_2\, (g)})$.