Question

Consider the following reaction: A ↔ B

Using the thermodynamic data provided below at 25 °C, determine the equilibrium constant for this reaction. R = 8.314 ^J/_{mol K} .

ΔG^o_f(A) = -2.2 kJ/mol

ΔG^o_f(B) = -4.8 kJ/mol

My key says the correct answer is 2.9, what do you use to get there?

Answer 1

$\Delta G_{reaction}=G_{B}-G_{A}$

$\Delta G_{reaction}=(-4.8+2.2)\: kJ/mol=-2.6\: kJ/mol$

:: AG reaction = (-4.8 +2.2) kJ/mol = -2600 J/mol

We know, the relatiobn between free energy and equilibrium constant-

$\Delta G=-RTlnK_{eq}$

-AG : In Keq = R XT

$\therefore K_{eq}=exp(\frac{-\Delta G}{R\times T})$

$\therefore K_{eq}=exp(\frac{-(-2600\: J/mol)}{8.314\: J/(mol.K)\times 298\: K})$

$\therefore K_{eq}=exp(1.049)$

$\therefore K_{eq}=2.859$

$\therefore K_{eq}\approx 2.9$ [ANSWER]