$$\ce{PCl3(g) + Cl2(g) <--> PCl5(g) } \hskip{2em} K_c=6.5 $$

At a certain point in time, a 1.00 L rigid reaction vessel contains 1.5 mol of \(\ce{PCl3(g)}\), 1.0 mol of \(\ce{Cl2(g)}\), and 2.5 mol of \(\ce{PCl5(g)}\). Which of the following describes how the measured pressure in the reaction vessel in the reaction vessel will change and why it will change that way as the reaction system approaches equilibrium at constant temperature?

Summary Submit

We can calculate the reaction quotient \(Q\):

$$ Q = \frac{[\ce{PCl5}]}{[\ce{PCl3}][\ce{Cl2}]} $$ $$ Q = \frac{(2.5)}{(1.5)(1)} $$ $$ Q = \frac{\frac{5}{2}}{\frac{3}{2}} $$ $$ Q = \frac{5}{2}\cdot \frac{2}{3} $$ $$ Q = \frac{5}{3} $$ $$ \frac{5}{3} \lt 6.5 $$ $$ Q \lt K_c $$

Since \(Q\) is less than \(K_c\), the equilibrium position will shift to the products. The pressure will decrease since two molecules of gas will turn into only 1 molecule of gas, resulting in a net decrease in gas pressure.