# How do you find equilibrium constant for a reversable reaction?

Mar 17, 2014

(a) Write the equilibrium expression for the reaction.
(b) Determine the molar concentrations or partial pressures of each species involved.
(c) Substitute into the equilibrium expression and solve for K.

1.Calculate the value of the equilibrium constant, Kc , for the system shown, if 0.1908 moles of C${O}_{2}$, 0.0908 moles of ${H}_{2}$, 0.0092 moles of CO, and 0.0092 moles of ${H}_{2}$O vapor were present in a 2.00 L reaction vessel were present at equilibrium.

C${O}_{2}$ (g) + ${H}_{2}$ (g) <-------> CO (g) + ${H}_{2}$O

(a) Write the equilibrium expression for the reaction.

${K}_{c}$ = [CO] [${H}_{2}$O] / [C${O}_{2}$] [ ${H}_{2}$O]

(b) Determine the molar concentrations or partial pressures of each species involved. ( the given equilibrium amounts are expressed in moles per liter )
[CO] = 0.0092 mol/ 2 L =0.0046 mol/L
[${H}_{2}$O] = 0.0908 mol/ 2 L =0.0454mol/L
[C${O}_{2}$] = 0.1908 mol / 2 L = 0.0954 mol/L
[ ${H}_{2}$O] =0.0092 mol/ 2 L =0.0046 mol/L

(c) Substitute into the equilibrium expression and solve for K.

${K}_{c}$ = [CO] [${H}_{2}$O] / [C${O}_{2}$] [ ${H}_{2}$O]

${K}_{c}$ = 00046 x 0.0046 / 0.0954 x 0.0454

${K}_{c}$ = 0.0049 or 4.9 x 1${0}^{-} 3$