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Chapter 15: Problem 31

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) .\) An equilibrium mixture in a 10.00-L vessel is found to contain \(0.050 \mathrm{~mol}\) \(\mathrm{CH}_{3} \mathrm{OH}, 0.850 \mathrm{~mol} \mathrm{CO},\) and \(0.750 \mathrm{~mol} \mathrm{H}_{2}\) at \(500 \mathrm{~K} .\) Calculate \(K_{c}\) at this temperature.

Short Answer

Expert verified
The equilibrium constant \(K_c\) at 500 K is approximately 10.458.

Step by step solution

01

Write the Expression for Kc

The equilibrium constant \(K_c\) for a reaction is expressed in terms of the concentrations of products and reactants raised to the power of their stoichiometric coefficients. For the reaction \( \text{CO}(g) + 2 \text{H}_2(g) \rightleftharpoons \text{CH}_3 \text{OH}(g) \), the expression is: \[ K_c = \frac{[\text{CH}_3 \text{OH}]}{[\text{CO}][\text{H}_2]^2} \]
02

Calculate Molar Concentrations

To find \(K_c\), we need the concentration of each species in molarity (mol/L). Given the volume of the container is 10.00 L:- \( [\text{CH}_3 \text{OH}] = \frac{0.050 \text{ mol}}{10.00 \text{ L}} = 0.005 \text{ M} \)- \( [\text{CO}] = \frac{0.850 \text{ mol}}{10.00 \text{ L}} = 0.085 \text{ M} \)- \( [\text{H}_2] = \frac{0.750 \text{ mol}}{10.00 \text{ L}} = 0.075 \text{ M} \)
03

Plug Values into the Kc Expression

Substitute the calculated concentrations into the expression for \(K_c\):\[ K_c = \frac{0.005}{(0.085)(0.075)^2} \]
04

Perform the Calculation

Calculate the expression:- Calculate \((0.075)^2 = 0.005625\)- Multiply \(0.085 \times 0.005625 = 0.000478125\)- Divide: \[ K_c = \frac{0.005}{0.000478125} \approx 10.458 \]
05

Conclusion

The equilibrium constant \(K_c\) is approximately 10.458 at 500 K.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
In the realm of chemical reactions, the equilibrium constant, denoted as \(K_c\), is a crucial measure. It tells us the ratio of the concentration of products to reactants at equilibrium.
At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, meaning the system's concentrations remain constant. This constancy is captured by \(K_c\), which depends on the reaction's specific stoichiometry.
  • To find \(K_c\), construct an expression involving the concentrations of the products and reactants, raised to the power of their coefficients in the balanced equation.
  • For the reaction \( ext{CO}(g) + 2 ext{H}_2(g) \rightleftharpoons ext{CH}_3 ext{OH}(g) \), the expression becomes: \[ K_c = \frac{[\text{CH}_3 \text{OH}]}{[\text{CO}][\text{H}_2]^2} \]
Knowing \(K_c\) helps predict the extent of a reaction: a large \(K_c\) suggests a higher concentration of products relative to reactants at equilibrium.
Reaction Stoichiometry
Stoichiometry is the backbone of chemistry, defining how reactants and products relate in a given reaction. In the case of our methanol production reaction, stoichiometry shows us that one molecule of carbon monoxide reacts with two molecules of hydrogen to produce one molecule of methanol.
This stoichiometry not only determines the balanced chemical equation but is also crucial when calculating equilibrium constants.
  • Coefficients in the balanced equation provide the powers to which concentration terms are raised in the \(K_c\) expression.
  • In our example, the coefficient for \( ext{H}_2 \) is 2, reflected as \([\text{H}_2]^2\) in the \(K_c\) expression.
Understanding stoichiometry ensures accurate calculations and insight into how changes in conditions might affect the reaction's yield.
Catalyzed Reactions
Reactions in industrial settings, like the production of methanol, often occur with the assistance of catalysts. A catalyst is a substance that speeds up a chemical reaction without undergoing permanent change itself.
In the catalyzed reaction of carbon monoxide and hydrogen to make methanol, a catalyst provides an alternate pathway for the reaction with a lower activation energy.
  • Though catalysts accelerate the achievement of equilibrium, they do not change the equilibrium position. Hence, \(K_c\) remains unaffected by the addition of a catalyst.
  • This property makes catalysts perfect for enhancing production efficiency without altering the fundamental dynamics of the chemical equation.
Leveraging catalysis in reactions allows industries to produce chemicals faster and more efficiently, which is essential for commercial applications.

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Most popular questions from this chapter

Consider the hypothetical reaction $$ \mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)+2 \mathrm{C}(g) $$ A flask is charged with \(100 \mathrm{kPa}\) of pure \(\mathrm{A}\), after which it is allowed to reach equilibrium at \(25^{\circ} \mathrm{C}\). At equilibrium, the partial pressure of \(\mathrm{B}\) is \(25 \mathrm{kPa} .(\mathbf{a})\) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p} ?\) (c) What could we do to maximize the yield of \(\mathrm{B}\) ?

A mixture of 0.140 mol of \(\mathrm{NO}, 0.060 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and 0.260 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 2.0-L vessel at \(330 \mathrm{~K}\). Assume that the following equilibrium is established: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ At equilibrium \(\left[\mathrm{H}_{2}\right]=0.010 \mathrm{M} .(\mathbf{a})\) Calculate the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O} .\) (b) Calculate \(K_{c}\).

The equilibrium constant for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ is \(K_{c}=1.3 \times 10^{-2}\) at \(1000 \mathrm{~K} .(\mathbf{a})\) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\). (c) Calculate \(K_{c}\) for \(\mathrm{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\).

At \(900^{\circ} \mathrm{C}, K_{p}=51.2\) for the equilibrium $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ If the pressure of \(\mathrm{NO}(g)\) is half the pressure of \(\mathrm{NOBr}(g)\), what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

The equilibrium constant \(K_{c}\) for \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons\) \(2 \mathrm{CO}(g)\) is 1.9 at \(1000 \mathrm{~K}\) and 0.133 at \(298 \mathrm{~K} .(\mathbf{a})\) If excess C is allowed to react with \(25.0 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) in a \(3.00-\mathrm{L}\) vessel at \(1000 \mathrm{~K}\), how many grams of CO are produced? (b) How many grams of \(\mathrm{C}\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of \(\mathrm{CO}\) be greater or smaller? (d) Is the reaction endothermic or exothermic?
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