## 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.

## Step-by-Step

## 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.