# Question #4c3b0

May 19, 2015

In your case, the reverse reaction will occur.

You know that you're dealing with an equilibrium reaction for which the equilibrium constant is equal to 0.154.

$\textcolor{red}{2} C {H}_{4 \left(g\right)} r i g h t \le f t h a r p \infty n s {C}_{2} {H}_{2 \left(g\right)} + \textcolor{b l u e}{3} {H}_{2 \left(g\right)}$

In order to determine what net reaction takes place when you add those quantities of methane, acetylene, and hydrogen to the reaction vessel, you need to figure out what the reaction quotient, ${Q}_{c}$, will be.

By comparing the value of the reaction quotient and the value of the equilibrium constant, you can determine which reaction, if any (meaning that you could already be at equilibrium), will be favored.

By definition, the reaction quotient is equal to

${Q}_{c} = \frac{{\left[{C}_{2} {H}_{2}\right]}_{0} \cdot {\left[{H}_{2}\right]}_{0}^{\textcolor{b l u e}{3}}}{{\left[C {H}_{4}\right]}_{0}^{\textcolor{red}{2}}}$

Determine the initial concentrations of the species involved in the reaction by using

$C = \frac{n}{V}$

${\left[C {H}_{4}\right]}_{0} = \text{7.00 moles"/"5.40 L" = "1.30 M}$

${\left[{C}_{2} {H}_{2}\right]}_{0} = \text{4.75 moles"/"5.40 L" = "0.880 M}$

${\left[{H}_{2}\right]}_{0} = \text{11.80 moles"/"5.40 L" = "2.19 M}$

This means that the value of ${Q}_{c}$ will be

${Q}_{c} = \frac{0.880 \cdot {2.19}^{\textcolor{b l u e}{3}}}{{1.80}^{\textcolor{red}{2}}} = \text{5.47}$

Notice that ${Q}_{c}$ is much bigger than ${K}_{c}$. This tells you that there are more products than reactants in the reaction vessel. To reduce the amount of products and increase the amount of the reactant, the equilibrium will shift to the left, favoring the reverse reaction

$2 C {H}_{4} \leftarrow {C}_{2} {H}_{2} + 3 {H}_{2}$