What is the equilibrium constant for a reaction at temperature 52.5 °C if the equilibrium constant at 21.3 °C is 42.7? For this reaction, ΔrH° = 10.4 kJ mol-1 .

Answer 1

Use the van't Hoff equation:

ln (K2 / K1) = -∆H/R (1/T2 - 1/T1)

K1 = 42.7

K2 = ?

T1 = 21.3º + 273 = 294.3

T2 = 52.5º + 273 = 325.5

R = 0.008314 kJ/Kmol

∆H = 10.4 kJ/mol

ln (K2/42.7) = -(10.4 / 0.008314) (1/325.5 - 1/294.3)

ln (K2/42.7) = -1251 (0.00307 - 0.00340)

ln (K2/42.7) = +0.410

K2/42.7 = 1.50

K2 = 64.3

(be sure to check the math)

Answer 2

Assuming that ΔH and ΔS are constant with T:

ln(K(T₂)/K₍T₁)) = -ΔH_rxn/R *( 1/T₂ - 1/T₁) (Take each side to the e power and multiply by K(T₁) to solve for K(T₂))

Make sure you use ΔH in J/mole and R = 8.314 J/mole-K and T's in Kelvin.