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 $_5$ .

Question

Equilibrium Equations: Calculating K
In the reaction of Cl[latex]_2[/latex] with PCl[latex]_3[/latex], the concentrations of reactants and products were determined experimentally at equilibrium and found to be 7.2 mol/L for PCl[latex]_3[/latex], 7.2 mol/L for Cl[latex]_2[/latex], and 0.050 mol/L for PCl[latex]_5[/latex].

PCl[latex]_3[/latex](g) + Cl[latex]_2[/latex](g) [latex] \rightleftarrows [/latex] PCl[latex]_5[/latex](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[latex]_5[/latex], divided by the product of the concentrations of the two reactants, PCl[latex]_3[/latex] and Cl[latex]_2[/latex]. 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.

Accepted Answer

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

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.

Chapter 7

Q. 7.9

Q. 7.9

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