Question

For the reaction

\[\ce{SrCO3(s) <=> SrO(s) + CO2(g)}\]

the value of equilibrium constant K_p = 2.2 × 10^-4 at 1002 K. Calculate K_c for the reaction.

Answer 1

For the reaction,

\[\ce{SrCO3(s) <=> SrO(s) + CO2(g)}\]

∆n_g = 1 – 0 = 1

∴ K_p = K_c (RT)

2.2 × 10^-4 = K_c (0.0821) (1002)

`"K"_"C" = (2.2 xx 10^-4)/(0.0821 xx 1002) = 2.674 xx 10^-6`