15.5: Calculating Equilibrium Constants - Chemistry LibreTexts

Solution

A The initial concentrations of the reactants are \([H_2]_i = [CO_2]_i = 0.0150\; M\). Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of \(H_2O\) as \(x\), then \(Δ[H_2O] = +x\). We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of \(x\). For example, 1 mol of \(CO\) is produced for every 1 mol of \(H_2O\), so the change in the \(CO\) concentration can be expressed as \(Δ[CO] = +x\). Similarly, for every 1 mol of \(H_2O\) produced, 1 mol each of \(H_2\) and \(CO_2\) are consumed, so the change in the concentration of the reactants is \(Δ[H_2] = Δ[CO_2] = −x\). We enter the values in the following table and calculate the final concentrations.

\[H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}\nonumber \]

ICE	\([H_2]\)	\([CO_2]\)	\([H_2O]\)	\([CO]\)
Initial	0.0150	0.0150	0	0
Change	\(−x\)	\(−x\)	\(+x\)	\(+x\)
Final	\((0.0150 − x)\)	\((0.0150 − x)\)	\(x\)	\(x\)

B We can now use the equilibrium equation and the given \(K\) to solve for \(x\):

\[K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(x)(x)}{(0.0150−x)(0.0150−x}=\dfrac{x^2}{(0.0150−x)^2}=0.106\nonumber \]

We could solve this equation with the quadratic formula, but it is far easier to solve for \(x\) by recognizing that the left side of the equation is a perfect square; that is,

\[\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106\nonumber \]

Taking the square root of the middle and right terms,

\[\dfrac{x}{(0.0150−x)} =(0.106)^{1/2}=0.326\nonumber \]

\[x =(0.326)(0.0150)−0.326x\nonumber \]

\[1.326x=0.00489\nonumber \]

\[x =0.00369=3.69 \times 10^{−3}\nonumber \]

C The final concentrations of all species in the reaction mixture are as follows:

• \([H_2]_f=[H_2]_i+Δ[H_2]=(0.0150−0.00369) \;M=0.0113\; M\)
• \([CO_2]_f =[CO_2]_i+Δ[CO_2]=(0.0150−0.00369)\; M=0.0113\; M\)
• \([H_2O]_f=[H_2O]_i+Δ[H_2O]=(0+0.00369) \;M=0.00369\; M\)
• \([CO]_f=[CO]_i+Δ[CO]=(0+0.00369)\; M=0.00369 \;M\)

We can check our work by inserting the calculated values back into the equilibrium constant expression:

\[K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.00369)^2}{(0.0113)^2}=0.107\nonumber \]

To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed.