# Question #7f43b

Oct 6, 2015

$\text{CH"_3"Cl}$ is being formed.

#### Explanation:

You know that the equilibrium constant, ${K}_{p}$, for this reaction is $1.6 \cdot {10}^{4}$ at $\text{1500 K}$, but you don't know how in which direction will the reaction proceed for those specific partial pressures.

To find out, you need to determine the reaction quotient, ${Q}_{p}$, which will tell you how the current partial pressures of the species that take part in the reaction compare with the equilibrium partial pressures.

In other words, the reaction quotient tells you what reaction is favored at these specific partial pressures in order for the equilibrium to be established.

The equilibrium reaction looks like this

${\text{CH"_text(4(g]) + "Cl"_text(2(g]) rightleftharpoons "CH"_3"Cl"_text((g]) + "HCl}}_{\textrm{\left(g\right]}}$

By definition, the equilibrium constant is equal to

${K}_{p} = \frac{\left({P}_{H C l}\right) \cdot \left({P}_{C {H}_{3} C l}\right)}{\left({P}_{C {H}_{4}}\right) \cdot \left({P}_{C {l}_{2}}\right)}$

The expression for ${K}_{p}$ uses the equilibrium partial pressures of the gases.

The reaction quotient, on the other hand, uses the partial pressures of the gases at a specific moment $i$

${Q}_{p} = \frac{{\left({P}_{H C l}\right)}_{i} \cdot {\left({P}_{C {H}_{3} C l}\right)}_{i}}{{\left({P}_{\textrm{C {H}_{4}}}\right)}_{i} \cdot {\left({P}_{C {l}_{2}}\right)}_{i}}$

Plug in you values and solve for ${Q}_{p}$

${Q}_{p} = \left(\left(0.47 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm")))) * (0.24color(red)(cancel(color(black)("atm")))))/( (0.13color(red)(cancel(color(black)("atm")))) * (0.035color(red)(cancel(color(black)("atm}}}}\right)\right) = 24.8$

This tells you that the reaction is not at equilibrium.

How does ${Q}_{p}$ compare with ${K}_{p}$?

It's clear that ${Q}_{p} < {K}_{p}$, which implies that you have too much of the reactants and not enough of the products. In other words, the pressure of the reactants is too high and the pressure of the products is too low.

To reach equilibrium, the forward reaction will be favored, meaning that the equilibrium will shift to the right.

If this is the case, then ${\text{CH}}_{4}$ will be consumed and $\text{CH"_3"Cl}$ will be formed.