Question

What is the value of for this aqueous reaction at 298 K? A+B↽⇀C+D ΔG°=12.86 kJ/mol K=

Answer 1

Answer:

Kc = 0.5951 (4 sig. figs.)

Explanation:

For A + B ⇄ C + D at standard thermodynamic conditions (298K, 1atm)

ΔG = ΔG° + R·T·lnQ => 0 = ΔG° + R·T·lnKc => ΔG° = - R·T·lnKc

=> lnKc = - ΔG°/R·T

ΔG° = +12.86 Kj/mol

R = 8.314 Kj/mol·K

T = 298K

lnKc = - (+12.86Kj) / (8.314Kj/mol·K)(298K) = - 0.519 mol⁻¹

Kc = e⁻⁰°⁵¹⁹ mol⁻¹ = 0.5957 mol⁻¹ (4 sig. figs.)