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Chapter 14: Problem 33

Methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) can be prepared in the presence of a catalyst by the reaction of \(\mathrm{H}_{2}\) and \(\mathrm{CO}\) at high temperatures according to the following equation: $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}(g)$$ What is the concentration of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in moles per liter if the concentration of \(\mathrm{H}_{2}=0.080 \mathrm{mol} / \mathrm{L}\) , the concentration of \(\mathrm{CO}=0.025 \mathrm{mol} / \mathrm{L},\) and \(K_{e q}=290\) at 700 \(\mathrm{K} ?\)

Short Answer

Expert verified
The concentration of \(\mathrm{CH}_{3}\mathrm{OH}\) is \(0.046 \, \mathrm{mol}/\mathrm{L}\).

Step by step solution

01

Write down the equilibrium expression

The equilibrium constant is given by the ratio of the concentrations of the products to the reactants, each raised to a power equal to the coefficient in the balanced equation. For the given reaction \(\mathrm{CO}(g) + 2\,\mathrm{H}_{2}(g) \rightleftarrows \(\mathrm{CH}_{3}\mathrm{OH}(g)\), the equilibrium expression is: \[ K_{eq} = \frac{[\mathrm{CH}_{3}\mathrm{OH}]}{[\mathrm{CO}]\cdot[\mathrm{H}_{2}]^{2}} \]
02

Set up the equation to solve for the concentration of \(\mathrm{CH}_{3}\mathrm{OH}\)

Plug in the given values of the equilibrium constant and the concentrations of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) into the equilibrium expression and solve for \([\mathrm{CH}_{3}\mathrm{OH}]\): \[ 290 = \frac{[\mathrm{CH}_{3}\mathrm{OH}]}{(0.025) \cdot (0.080)^{2}} \]
03

Solve for the concentration of \(\mathrm{CH}_{3}\mathrm{OH}\)

Rearrange the equation and solve for \([\mathrm{CH}_{3}\mathrm{OH}]\): \[ [\mathrm{CH}_{3}\mathrm{OH}] = 290 \cdot (0.025) \cdot (0.080)^{2} \]
04

Calculate and round the final result

After calculation, round the result to two significant figures, as the given values had only two significant figures: \[ [\mathrm{CH}_{3}\mathrm{OH}] = 0.046 \, \mathrm{mol}/\mathrm{L} \]

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Most popular questions from this chapter

Vinegar-a solution of acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) in water \(-\) is used in varying concentrations for different household tasks. The following equilibrum exists in vinegar. $$\begin{array}{c}{\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows} \\ {\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)}\end{array}$$ If the concentration of the acetic acid solution at equilibrium is 3.00 \(\mathrm{M}\) and the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=7.22 \times 10^{-3},\) what is the \(K_{e q}\) value for acetic acid?

What is the solubility product for copper \((\mathrm{I})\) sulfide, \(\mathrm{Cu}_{2} \mathrm{S},\) given that the solubility of \(\mathrm{Cu}_{2} \mathrm{S}\) is \(8.5 \times 10^{-17} \mathrm{M} ?\)

Does the reverse reaction rate ever equal zero? Why or why not?

What is the concentration of \(\mathrm{F}^{-}\) ions in a saturated solution that is 0.10 \(\mathrm{M}\) in \(\mathrm{Ca}^{2+}\) ? The \(K_{s p}\) of \(\mathrm{CaF}_{2}\) is \(1.6 \times 10^{-10} .\)

In the context of Le Chatelier's principle, what is a "stress"? List the stresses that affect chemical equilibria.
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