Chapter 7

Q. 7.9

Equilibrium Equations: Calculating K
In the reaction of Cl_2 with PCl_3, the concentrations of reactants and products were determined experimentally at equilibrium and found to be 7.2 mol/L for PCl_3, 7.2 mol/L for Cl_2, and 0.050 mol/L for PCl_5.

PCl_3(g) + Cl_2(g) \rightleftarrows PCl_5(g)

Write the equilibrium equation, and calculate the equilibrium constant for the reaction. Which reaction is favored, the forward one or the reverse one?

ANALYSIS All the coefficients in the balanced equation are 1, so the equilibrium constant equals the concentration of the product, PCl_5, divided by the product of the concentrations of the two reactants, PCl_3 and Cl_2. Insert the values given for each concentration, and calculate the value of K.

BALLPARK ESTIMATE At equilibrium, the concentration of the reactants (7.2 mol/L for each reactant) is higher than the concentration of the product (0.05 mol/L), so we expect a value of K less than 1.


Verified Solution

K = \frac{[PCl_5 ]}{[PCl_3 ] [Cl_2 ]} = \frac{0.050\ mol/L}{(7.2\ mol/L)(7.2\ mol/L)} = 9.6 × 10^{-4}

The value of K is less than 1, so the reverse reaction is favored. Note that units for K are omitted.

BALLPARK CHECK Our calculated value of K is just as we predicted: K < 1.