1. For the reaction, #PCl_(3(g))+Cl_(2(g)) rightleftharpoons PCl_(5(g))#, #K_c = 96.2 # at 400 K. If the initial concentrations are 0.22 mol/L of #PCl_3# and 0.42 mol/L of #Cl_2#, what are the equilibrium concentrations of all species?

Question

Olivia Anton · Accepted Answer

You are aware that, at a given temperature, the equilibrium constant of the reaction you are working with is 96.2. The fact that #K_c# is larger than one tells you that the reaction will favor the product, #PCl_5#, at equilibrium. Moreover, since you only start with reactants, you can predict that the concentrations of both #PCl_3#, and of #Cl_2# will decrease. At the same time, the concentration of #PCl_5# will increase. Use an ICE table to help you determine the equilibrium concentrations for your reaction #" "PCl_(3(s)) + Cl_(2(g)) rightleftharpoons PCl_(5(g))#
I.....0.22...........0.42.................0
C.....(-x).............(-x)..................(+x)
E...0.22-x.......0.42-x................x The equilibrium constant will, by definition, be equal to #K_c = ([PCl_5])/([PCl_3] * [Cl_2]) = x/((0.22-x) * (0.42 - x)) = 96.2# Put this equation back in quadratic form. #96.2 * (0.22-x) * (0.42-x) = x# #96.2 * (0.0924 - 0.22x - 0.42x + x^2) = x# #96.2x^2 - 62.568x + 8.888 = 0# This equation will produce two solutions for #x# #{ (cancel(x_1 = 0.4408)), (x_2 = 0.2096) :}# The first solution is eliminated because it will result in negative equilibrium concentrations for #PCl_3# and #Cl_2#, which means that you'll get  #[PCl_3] = 0.22 - 0.2096 = color(green)("0.0104 M")#
#[Cl_2] = 0.42 - 0.2096 = color(green)("0.210 M")#
#[PCl_5] = 0 + 0.2096 = color(green)("0.210 M")# SIDE NOTE: I've included three sig figs with the answers.

Luna Chen · Answer

undefined To find the equilibrium concentrations of all species, we'll use the initial concentrations and the equilibrium constant ($ K_c $).

Given:
Initial concentration of $ PCl_3 $ ($ [PCl_3]_0 $) = 0.22 mol/L
Initial concentration of $ Cl_2 $ ($ [Cl_2]_0 $) = 0.42 mol/L
Equilibrium constant ($ K_c $) = 96.2

The balanced equation for the reaction is:

$$ PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) $$

Let's assume that at equilibrium, the concentrations of $ PCl_3 $, $ Cl_2 $, and $ PCl_5 $ are $ [PCl_3]_{eq} $, $ [Cl_2]_{eq} $, and $ [PCl_5]_{eq} $, respectively.

The expression for the equilibrium constant ($ K_c $) is:

$$ K_c = \frac{[PCl_5]_{eq}}{[PCl_3]_{eq} \times [Cl_2]_{eq}} $$

Given that $ K_c = 96.2 $, we can set up the equation as:

$$ 96.2 = \frac{[PCl_5]_{eq}}{[PCl_3]_{eq} \times [Cl_2]_{eq}} $$

We know that $ [PCl_3]_{eq} = [PCl_3]_0 - x $ and $ [Cl_2]_{eq} = [Cl_2]_0 - x $, where $ x $ is the change in concentration at equilibrium. Since 1 mol of $ PCl_3 $ reacts with 1 mol of $ Cl_2 $ to form 1 mol of $ PCl_5 $, the change in concentration of $ PCl_3 $ and $ Cl_2 $ is equal to $ x $, while the change in concentration of $ PCl_5 $ is equal to $ +x $.

Substituting the given values into the equation and solving for $ x $, we can determine the equilibrium concentrations of all species.