Questions: Consider the following reaction: CO(g) + 2 H2(g) = CH3OH(g) The reaction between CO and H2 is carried out at a specific temperature with initial concentrations of CO=0.32 M and H2=0.48 M. At equilibrium, the concentration of CH3OH is 0.11 M Find the equilibrium constant at this temperature. Express your answer using two significant figures. Kc=

Solution Steps

For the given reaction:

\[ \mathrm{CO}(\mathrm{~g}) + 2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{~g}) \]

The equilibrium constant expression \( K_c \) is given by:

\[ K_c = \frac{[\mathrm{CH}_3\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_2]^2} \]

We are given the initial concentrations and the equilibrium concentration of \(\mathrm{CH}_3\mathrm{OH}\):

• Initial \([\mathrm{CO}] = 0.32 \, \mathrm{M}\)
• Initial \([\mathrm{H}_2] = 0.48 \, \mathrm{M}\)
• Equilibrium \([\mathrm{CH}_3\mathrm{OH}] = 0.11 \, \mathrm{M}\)

Assuming \(x\) moles of \(\mathrm{CH}_3\mathrm{OH}\) are formed, the change in concentration for \(\mathrm{CO}\) and \(\mathrm{H}_2\) will be:

• \([\mathrm{CO}] = 0.32 - x\)
• \([\mathrm{H}_2] = 0.48 - 2x\)

Since \([\mathrm{CH}_3\mathrm{OH}] = 0.11 \, \mathrm{M}\), we have \(x = 0.11\).

Substitute \(x = 0.11\) into the expressions for \([\mathrm{CO}]\) and \([\mathrm{H}_2]\):

• \([\mathrm{CO}] = 0.32 - 0.11 = 0.21 \, \mathrm{M}\)
• \([\mathrm{H}_2] = 0.48 - 2(0.11) = 0.26 \, \mathrm{M}\)

Substitute the equilibrium concentrations into the \(K_c\) expression:

\[ K_c = \frac{0.11}{0.21 \times (0.26)^2} \]

Calculate:

\[ K_c = \frac{0.11}{0.21 \times 0.0676} = \frac{0.11}{0.014196} \approx 7.75 \]

Final Answer