Question

carbon dioxide and water react to form methanol and oxygen, like this:
2co₂(g)+4h₂o(g)→2ch₃oh(l)+3o₂(g)
at a certain temperature, a chemist finds that a 9.0 l reaction vessel containing a mixture of carbon dioxide, water, methanol, and oxygen at equilibrium has the following composition:
calculate the value of the equilibrium constant k_c for this reaction. round your answer to 2 significant digits.
k_c =

Explanation:

Step1: Calculate the molarity of each gas

• For \(CO_2\):

The molar mass of \(CO_2\) is \(M_{CO_2}=44.01\space g/mol\).
The number of moles \(n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{2.84\space g}{44.01\space g/mol}\approx0.0645\space mol\).
Molarity \( [CO_2]=\frac{n_{CO_2}}{V}=\frac{0.0645\space mol}{9.0\space L}\approx7.17\times10^{-3}\space M\).

• For \(H_2O\):

The molar mass of \(H_2O\) is \(M_{H_2O} = 18.02\space g/mol\).
The number of moles \(n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{4.74\space g}{18.02\space g/mol}\approx0.263\space mol\).
Molarity \( [H_2O]=\frac{n_{H_2O}}{V}=\frac{0.263\space mol}{9.0\space L}\approx2.92\times10^{-2}\space M\).

• For \(O_2\):

The molar mass of \(O_2\) is \(M_{O_2}=32.00\space g/mol\).
The number of moles \(n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{2.44\space g}{32.00\space g/mol}\approx0.0763\space mol\).
Molarity \( [O_2]=\frac{n_{O_2}}{V}=\frac{0.0763\space mol}{9.0\space L}\approx8.48\times10^{-3}\space M\).
Methanol (\(CH_3OH\)) is a liquid and is not included in the \(K_c\) expression for gas - phase reactions.

Step2: Write the \(K_c\) expression

The balanced chemical equation is \(2CO_2(g)+4H_2O(g)
ightleftharpoons2CH_3OH(l)+3O_2(g)\).
The equilibrium constant expression \(K_c=\frac{[O_2]^3}{[CO_2]^2[H_2O]^4}\).

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

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Answer:

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