The equilibrium constant (Kc) for the reaction is given by the expression:
Kc = [H2][F2]/[HF]^2
Given that the initial concentration of HF is 0.025M and the equilibrium concentration of H2 is 0.0028M, we can express this as:
[H2] = 0.0028M
[F2] = 0.0028M (since the reaction stoichiometry indicates that the concentration of H2 is the same as F2)
[HF] = 0.025M - 2(0.0028M) = 0.0194M
Substitute these values into the expression for Kc:
Kc = (0.0028)(0.0028)/(0.0194)^2
Kc = 0.00000784/0.037636
Kc = 0.000208
Therefore, the equilibrium constant (Kc) for the reaction is 0.000208.
Given the reaction:
2HF (g) = H2 (g)+ F2 (g) if the initial concentration of. HF is 0.025M and the equilibrium concentration of H2 is 0.0028M, Then what is the equilibrium constant of the reaction?
1 answer