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

Choice A:
`5.32\; \rm kJ \cdot mol^(-1)`.

Step-by-step 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 * 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)})`.