Consider the following reaction between methanol and carbon monoxide to generate acetic acid:𝐶𝐻3𝑂𝐻+𝐶𝑂⇌𝐶𝐻3𝐶𝑂𝑂𝐻 𝐾𝑐=0.27This reaction is run for 4 hours, after which it has reached a steady state. If the concentration of methanol is 1.39 M and the concentration of carbon monoxide is 1.67 M, what is the concentration of acetic acid to 2 decimal places?
Question
Consider the following reaction between methanol and carbon monoxide to generate acetic acid:𝐶𝐻3𝑂𝐻+𝐶𝑂⇌𝐶𝐻3𝐶𝑂𝑂𝐻 𝐾𝑐=0.27This reaction is run for 4 hours, after which it has reached a steady state. If the concentration of methanol is 1.39 M and the concentration of carbon monoxide is 1.67 M, what is the concentration of acetic acid to 2 decimal places?
Solution
The reaction given is:
CH3OH + CO ⇌ CH3COOH
The equilibrium constant expression for this reaction is:
Kc = [CH3COOH] / ([CH3OH][CO])
We are given that Kc = 0.27, [CH3OH] = 1.39 M, and [CO] = 1.67 M. We are asked to solve for [CH3COOH].
Substituting the given values into the equilibrium constant expression gives:
0.27 = [CH3COOH] / (1.39 * 1.67)
Solving this equation for [CH3COOH] gives:
[CH3COOH] = 0.27 * 1.39 * 1.67 = 0.62 M
So, the concentration of acetic acid at equilibrium is 0.62 M to two decimal places.
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