# During the course of the reaction, the vessel is found to contain "7.00 mols" of C, "14.0 mols" of H_2O, "3.60 mols" of CO, and "8.50 mols" of H_2. What is the reaction quotient Q?

## The following reaction was carried out in a $\text{2.25 L}$ reaction vessel at $\text{1100 K}$: $C \left(s\right) + {H}_{2} O \left(g\right) r i g h t \le f t h a r p \infty n s C O \left(g\right) + {H}_{2} \left(g\right)$

Jul 23, 2016

${Q}_{c} = 0.972$

#### Explanation:

The equilibrium given to you looks like this

${\text{C"_ ((s)) + "H"_ 2"O"_ ((g)) rightleftharpoons "CO"_ ((g)) + "H}}_{2 \left(g\right)}$

By definition, the reaction quotient, ${Q}_{c}$, tells you the ratio that exists between the concentrations of the products raised to the power of their respective stoichiometric coefficients and the concentrations of the reactants raised to the power of their respective stoichiometric coefficients.

Two important things to note here

• the reaction quotient can be calculated by using the concentrations of the chemical species that take part in the reaction at any given moment in the course of the reaction
• the concentration of solids and of pure liquids is excluded from the expression of the reaction quotient

In this case, the reaction quotient would take the form -- keep in mind that carbon, $\text{C}$, is in the solid state in this reaction

${Q}_{c} = \left(\left[\text{CO"] * ["H"_2])/(["H"_2"O}\right]\right)$

Now, use the volume of the reaction vessel to calculate the concentrations of the chemical species that are of interest here

color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/(V_"solution"[color(blue)("in liters")])color(white)(a/a)|)))

You will have

["H"_2"O"] = "14.0 moles"/"2.25 L" = "6.22 M"

["CO"] = "3.60 moles"/"2.25 L" = "1.6 M"

["H"_2] = "8.50 moles"/"2.25 L" = "3.78 M"

Plug these values into the equation that gives you the reaction quotient to find

Q_c = ("1.6 M" * 3.78 color(red)(cancel(color(black)("M"))))/(6.22color(red)(cancel(color(black)("M")))) = "0.972 M"

${Q}_{c} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{0.972} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.