Question

methanol and oxygen react to form carbon dioxide and water, like this: 2ch₃oh(l)+3o₂(g)→2co₂(g)+4h₂o(g) at a certain temperature, a chemist finds that a 6.9 l reaction vessel containing a mixture of methanol, oxygen, carbon dioxide, and water at equilibrium has the following composition: calculate the value of the equilibrium constant kₑ for this reaction. round your answer to 2 significant digits. kₑ =

Explanation:

Step1: Calculate molarity of each compound

• Molar mass of \(CH_3OH\): \(M_{CH_3OH}=(12 + 4\times1+16)\space g/mol = 32\space g/mol\). Molarity \(c_{CH_3OH}=\frac{n}{V}=\frac{\frac{3.53\space g}{32\space g/mol}}{6.9\space L}\approx1.61\times10^{-2}\space M\)
• Molar mass of \(O_2\): \(M_{O_2}=32\space g/mol\). Molarity \(c_{O_2}=\frac{n}{V}=\frac{\frac{2.12\space g}{32\space g/mol}}{6.9\space L}\approx9.63\times10^{-3}\space M\)
• Molar mass of \(CO_2\): \(M_{CO_2}=44\space g/mol\). Molarity \(c_{CO_2}=\frac{n}{V}=\frac{\frac{2.32\space g}{44\space g/mol}}{6.9\space L}\approx7.63\times10^{-3}\space M\)
• Molar mass of \(H_2O\): \(M_{H_2O}=18\space g/mol\). Molarity \(c_{H_2O}=\frac{n}{V}=\frac{\frac{4.91\space g}{18\space g/mol}}{6.9\space L}\approx3.96\times10^{-2}\space M\)

Step2: Write the equilibrium constant expression

For the reaction \(2CH_3OH(l)+3O_2(g)
ightleftharpoons2CO_2(g)+4H_2O(g)\), the equilibrium constant \(K_c=\frac{[CO_2]^2[H_2O]^4}{[O_2]^3}\) (liquid \(CH_3OH\) is excluded from the expression as its concentration is constant in a reaction mixture where it is a pure liquid)

Step3: Substitute molarity values into \(K_c\) expression

\(K_c=\frac{(7.63\times 10^{-3})^2\times(3.96\times 10^{-2})^4}{(9.63\times 10^{-3})^3}\)

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Answer:

\(1.6\times 10^{-5}\)