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A mixture of carbon monoxide, hydrogen, and methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) is at equilibrium according to the equation $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ At \(290^{\circ} \mathrm{C},\) the mixture is \(0.034 \mathrm{M} \mathrm{CO}, 0.450 \mathrm{M} \mathrm{H}_{2},\) and \(0.00023 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\). What is \(K_{c}\) for this reaction at \(290^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The equilibrium constant \( K_c \) is approximately 0.0334.

Step by step solution

01

Understand the Equilibrium Expression

For the given reaction \( \text{CO}(g) + 2 \text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g) \), the equilibrium constant \( K_c \) is expressed as \( K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2} \). This formula relates the concentrations of reactants and products at equilibrium.
02

Substitute Given Values into the Expression

Substitute the equilibrium concentrations into the expression: \[ K_c = \frac{0.00023}{0.034 \times (0.450)^2} \]. This step places all given concentration values in their respective spots according to the equilibrium expression.
03

Calculate the Denominator

Calculate \( [\text{CO}][\text{H}_2]^2 \) by multiplying the concentration of CO (0.034 M) with the square of the concentration of \( \text{H}_2 \) (0.450 M): \( [\text{CO}][\text{H}_2]^2 = 0.034 \times 0.450^2 \). First, calculate \( 0.450^2 = 0.2025 \) and then \( 0.034 \times 0.2025 = 0.006885 \).
04

Calculate \( K_c \)

Using the calculated denominator, find \( K_c \) by dividing the concentration of methanol \( [\text{CH}_3\text{OH}] = 0.00023 \) by the calculated denominator value: \[ K_c = \frac{0.00023}{0.006885} \approx 0.0334 \].
05

Conclusion

After performing the division, we find that the equilibrium constant \( K_c \) for this reaction at \( 290^{\circ} \text{C} \) is approximately 0.0334.

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Key Concepts

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

Equilibrium Constant (Kc)
The equilibrium constant, denoted as \( K_c \), is a fundamental concept in chemical equilibrium. It provides a snapshot of the relative concentrations of reactants and products in a reaction at equilibrium. For a general reaction, the expression for \( K_c \) is constructed using the concentrations of the products raised to the power of their stoichiometric coefficients, divided by the concentrations of the reactants raised likewise to their coefficients. This ratio remains constant at a given temperature. In the case of the reaction: \( \text{CO}(g) + 2 \text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g) \), the equilibrium expression is given by:
  • \( K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2} \)
This formula reflects the balance of forward and reverse reactions. A high \( K_c \) indicates that the reaction yields more products at equilibrium, while a low \( K_c \) suggests that reactants dominate in the mixture. In our calculation exercise, substituting the concentrations leads to the approximation of \( K_c \) as 0.0334 at 290°C.
Reaction Quotient
The reaction quotient, denoted as \( Q \), serves a similar function as \( K_c \) but can be calculated at any point during a reaction, not just at equilibrium. While \( K_c \) is a constant at a given temperature, \( Q \) helps in determining the direction a reaction will proceed to reach equilibrium. It's calculated using the same expression as \( K_c \), involving the concentrations of products and reactants. However, the key is:
  • If \( Q < K_c \), the reaction will proceed in the forward direction to produce more products.
  • If \( Q > K_c \), the reaction will shift in the reverse direction, forming more reactants.
  • If \( Q = K_c \), the system is at equilibrium.
Calculating \( Q \) at different stages allows chemists to predict and control reactions to obtain desired products more efficiently.
Le Chatelier's Principle
Le Chatelier's Principle is a powerful tool in understanding how changes in conditions affect the chemical equilibrium of a system. The principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and re-establish equilibrium. Here are some of the factors affecting equilibrium:
  • Concentration changes: Adding or removing substances will shift the equilibrium to oppose the change.
  • Temperature changes: Increasing temperature favors the endothermic direction of a reaction, while decreasing favors the exothermic side.
  • Pressure changes: For gaseous reactions, increasing pressure shifts equilibrium to the side with fewer moles of gas; decreasing pressure favors the side with more gas.
  • Catalysts: They do not affect the position of equilibrium but can speed up the rate at which equilibrium is achieved.
By applying Le Chatelier's Principle, chemists can predict and manipulate the outcome of reactions to synthesize desired chemicals or maximize yield.

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Most popular questions from this chapter

The equilibrium-constant expression for a gas reaction is $$ K_{c}=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}\left[\mathrm{SO}_{2}\right]^{2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}\left[\mathrm{O}_{2}\right]^{3}} $$ Write the balanced chemical equation corresponding to this expression.

One way of preparing hydrogen is by the decomposition of water. $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) ; \Delta H^{\circ}=484 \mathrm{~kJ} $$ Would you expect the decomposition to be favorable at high or low temperature? Explain.

Nitrogen monoxide, \(\mathrm{NO},\) is formed in automobile exhaust by the reaction of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) (from air). $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ The equilibrium constant \(K_{c}\) is 0.0025 at \(2127^{\circ} \mathrm{C}\). If an equilibrium mixture at this temperature contains \(0.023 \mathrm{~mol} \mathrm{~N}_{2}\) and \(0.031 \mathrm{~mol} \mathrm{O}_{2}\) per liter, what is the concentration of \(\mathrm{NO}\) ?

The equilibrium constant \(K_{c}\) for the equation $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ is 1.84 at \(425^{\circ} \mathrm{C}\). The equilibrium constant \(K_{c}\) for the equation $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ is 49.7 at \(458^{\circ} \mathrm{C}\). Is this second equation endothermic or exothermic? How do you know? Describe the effect on this second equation of an increase in temperature. What would be the effect of an increase in pressure?

Methanol is prepared industrially from synthesis gas \(\left(\mathrm{CO}\right.\) and \(\left.\mathrm{H}_{2}\right)\) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) ; \Delta H^{\circ}=-21.7 \mathrm{kcal}\) Would the fraction of methanol obtained at equilibrium be increased by raising the temperature? Explain.

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