The ratio K_P/K_C for the reaction :
CO_((g))+1/2O_(2( g))⇌ CO_(2( g)) is :

Solution

Chemistry ma’am · 3 steps

Everything rests on Δ n_g: count gas moles on each side, and keep that half mole of oxygen.

1. The link. For any gaseous equilibrium, K_P and K_C are connected through Δ n_g, the moles of gaseous products minus the moles of gaseous reactants.

   K_P = K_C (RT)^(Δ n_g) ⇒ K_P/K_C = (RT)^(Δ n_g)

2. Find Δn. Products: 1 mol of CO₂. Reactants: 1 mol of CO plus 1/2 mol of O₂, so 3/2 mol. So Δ n_g = 1 - 3/2 = -1/2.
3. So. K_P/K_C = (RT)^(-1/2) = 1/√RT, which is option (d).

The idea to keep. Always take Δ n_g as gaseous products minus gaseous reactants, fractions included. A negative Δ n_g puts RT in the denominator.

Answer from the NTA final answer key. Solution by PrepWiser.