Question

Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the system?
CuS(s) + O2(g) ? Cu(s) + SO2(g)
Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the system?
CuS(s) + O2(g) ? Cu(s) + SO2(g)
a) The equilibrium constant will increase.
b) The reaction will shift to the left in the direction of reactants.
c) The reaction will shift to the right in the direction of products.
d) No effect will be observed.
e) The equilibrium constant will decrease.

Answer 1

Answer:

Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.

Explanation:

The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:

$Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}$.

Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both $[\mathrm{SO_2\, (g)}]$ and $[\mathrm{O_2\, (g)}]$ will increase if the pressure is increased through compression. However, because $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient $Q$.

As a result, the increase in pressure will have no impact on the value of $Q\!$. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.