Question

CO(g)+2H2(g)⇌CH3OH(g) A reaction mixture in a 5.25 −L flask at a certain temperature initially contains 27.1 g CO and 2.34 g H2. At equilibrium, the flask contains 8.65 g CH3OH.

Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Answer 1

               CO(g) +        2H2(g)          <--->              CH3OH(g)

initial      0.185                0.223                           0

at equilibrium 0.185-0.0514 = 0.1336 M 0.223-0.0514 = 0.1716 M       0.0514

Initial

No of moles of CO = 27.1/28 = 0.97 mol

concentration of CO = 0.97/5.25 = 0.185 M

No of moles of H2 = 2.34/2 = 1.17 mol

concentration of H2 = 1.17/5.25 = 0.223 M

atequilibrium

No of moles of CH3OH = 8.65/32 = 0.27 mol

concentration of CH3OH = 0.27/5.25 = 0.0514 M

Kc = [CH3OH]/[H2]^2[CO]

   = 0.0514/(0.223^2*0.185)

     = 5.6