Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position
Problem How would you change the volume for each of the following reactions to increase the yield of the product(s)?
(a) CaCO_3(s) \xrightleftharpoons CaO(s) + CO_2(g) (b) S(s) + 3F_2(g) \xrightleftharpoons SF_6(g)
(c) Cl_2(g) + I_2(g) \xrightleftharpoons 2ICl(g)
Plan Whenever gases are present, a change in volume causes a change in concentration. For reactions in which the number of moles of gas changes, if the volume decreases (pressure increases), the equilibrium position shifts to lower the pressure by reducing the number of moles of gas. A volume increase (pressure decrease) has the opposite effect.
Solution (a) The only gas is the product CO_2. To make the system produce more molecules of gas, that is, more CO_2, we increase the volume (decrease the pressure).
(b) With 3 mol of gas on the left and only 1 mol on the right, we decrease the volume (increase the pressure) to form fewer molecules of gas and, thus, more SF_6.
(c) The number of moles of gas is the same on both sides of the equation, so a change in volume (pressure) will have no effect on the yield of ICl.
Check Let’s predict the relative values of Q_c and K_c.
(a) Q_c = [CO_2], so increasing the volume will make Q_c < K_c, and the system will yield more CO_2.
(b) Q_c = [SF_6]/[F_2]^3. Lowering the volume increases [F_2] and [SF_6] proportionately, but Q_c decreases because of the exponent 3 in the denominator. To make Q_c = K_c again, [SF_6] must increase.
(c) Q_c = [ICl]^2/[Cl_2][I_2]. A change in volume (pressure) affects the numerator (2 mol) and denominator (2 mol) equally, so it will have no effect.