The following reaction has attained equilibrium CO (g) + 2H2 (g) ⇌ CH3OH (g); ΔH° = – 92.0 kJ mol-1. What will happen if (i) the volume of the vessel is suddenly reduced to half?
Kc = [CH3OH]/[CO][H2]2, Kp = PCH3OH/PCO × PH2 When the volume of the vessel is reduced to half, the concentration of each reactant or product becomes double. Thus Qc = 2[CH3OH]/2[CO] × {2[H2]}2 = i K.. As Qc < Kp, equilibrium will shift in the forward direction.
