/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Methanol, \(\mathrm{CH}_{3} \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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Methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is manufactured industrially by the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ A gaseous mixture at \(500 \mathrm{~K}\) is \(0.020 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}, 0.10 \mathrm{M} \mathrm{CO}\), and \(0.10 M \mathrm{H}_{2} .\) What will be the direction of reaction if this mixture goes to equilibrium? The equilibrium constant \(K_{c}\) equals \(10.5\) at \(500 \mathrm{~K}\).

Short Answer

Expert verified
The reaction will shift to the left.

Step by step solution

01

Write the Reaction Expression for Checking Equilibrium

The given chemical reaction is \(\mathrm{CO}(g) + 2\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3}\mathrm{OH}(g)\). For this reaction, the equilibrium expression is:\[K_c = \frac{[\mathrm{CH}_{3}\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_{2}]^2}\]
02

Calculate the Current Reaction Quotient

Plug the given concentrations into the expression for the reaction quotient \(Q_c\):\[Q_c = \frac{[\mathrm{CH}_{3}\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_{2}]^2} = \frac{0.020}{0.10 \times (0.10)^2} = \frac{0.020}{0.001} = 20\]
03

Compare the Reaction Quotient to the Equilibrium Constant

We obtained \(Q_c = 20\), and the equilibrium constant \(K_c = 10.5\). To predict the direction of the reaction, compare these values:- If \(Q_c > K_c\), the reaction will shift to the left.- If \(Q_c < K_c\), the reaction will shift to the right.- If \(Q_c = K_c\), the system is already at equilibrium.

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Key Concepts

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

Understanding Reaction Quotient
The reaction quotient, represented as \(Q_c\), is a valuable tool in understanding the current state of a chemical reaction relative to its equilibrium. It is similar to the equilibrium constant \(K_c\), but \(Q_c\) is computed using the initial concentrations or partial pressures of the reactants and products at any given moment. To calculate \(Q_c\), we use the same expression as \(K_c\), which in this case is
  • \(Q_c = \frac{[\mathrm{CH}_{3}\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_{2}]^2}\)
By substituting the initial concentrations into this expression, we assess the current status of the reaction. This calculation does not indicate whether the reaction is at equilibrium, but it helps us determine how the reaction is progressing. Understanding \(Q_c\) is crucial for managing reactions in industrial settings, such as methanol production, by adjusting conditions to achieve the desired outcome.
Significance of the Equilibrium Constant
The equilibrium constant, \(K_c\), is an intrinsic property of a chemical reaction at a given temperature. It denotes the ratio of product concentrations to reactant concentrations when the reaction has reached equilibrium. For the methanol formation reaction,
  • \(K_c = 10.5\) at 500\(\,\mathrm{K}\)
This value reflects the balance between methanol and its reactants under standard conditions. At equilibrium, the reaction has no net change in reactant and product concentrations, though both forward and reverse reactions continue to occur at the same rate.
The constant provides insight into the favorability of the products over reactants. A large \(K_c\) value implies product favorability, while a small \(K_c\) indicates favorability towards reactants. Being aware of \(K_c\) allows chemists to predict the extent of a reaction and make adjustments to optimize yield in processes like the industrial production of methanol.
Predicting Reaction Direction
Predicting which direction a reaction will proceed is essential for controlling chemical processes. By comparing the reaction quotient \(Q_c\) to the equilibrium constant \(K_c\), we can predict the shift direction:
  • If \(Q_c > K_c\), the reaction proceeds to the left, favoring the formation of reactants.
  • If \(Q_c < K_c\), the reaction shifts to the right, producing more products.
  • When \(Q_c = K_c\), the system is at equilibrium, and no net change occurs.
In the methanol production scenario, with \(Q_c = 20\) and \(K_c = 10.5\), we see that \(Q_c\) exceeds \(K_c\). Therefore, the reaction will shift to the left, consuming methanol to form more carbon monoxide and hydrogen until equilibrium is reached. Recognizing these conditions helps chemists make strategic adjustments to reaction conditions, such as temperature and concentration, to steer the reaction towards the desired product effectively.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\) in a closed system, ammonium hydrogen sulfide exists as the following equilibrium: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ a. When a sample of pure \(\mathrm{NH}_{4} \mathrm{HS}(s)\) is placed in an evacuated reaction vessel and allowed to come to equilibrium at \(25^{\circ} \mathrm{C}\), total pressure is \(0.660 \mathrm{~atm} .\) What is the value of \(K_{p} ?\) b. To this system, sufficient \(\mathrm{H}_{2} \mathrm{~S}(g)\) is injected until the pressure of \(\mathrm{H}_{2} \mathrm{~S}\) is three times that of the ammonia at equilibrium. What are the partial pressures of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) ? C. In a different experiment, \(0.750\) atm of \(\mathrm{NH}_{3}\) and \(0.500\) atm of \(\mathrm{H}_{2} \mathrm{~S}\) are introduced into a \(1.00-\mathrm{L}\) vessel at \(25^{\circ} \mathrm{C}\). How many moles of \(\mathrm{NH}_{4} \mathrm{HS}\) are present when equilibrium is established?

List four ways in which the yield of ammonia in the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; \Delta H^{\circ}<0 $$ can be improved for a given amount of \(\mathrm{H}_{2}\). Explain the principle behind each way.

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

On the basis of the value of \(K_{c}\), decide whether or not you expect nearly complete reaction at equilibrium for each of the following: a. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) ; K_{c}=4.6 \times 10^{-31}\) b. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(g) ; K_{c}=1.3 \times 10^{21}\)

Part 1: You run the chemical reaction \(\mathrm{C}(a q)+\mathrm{D}(a q) \rightleftharpoons\) \(2 \mathrm{E}(a q)\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant \(K_{c}\) for the reaction at this temperature is \(2.0 .\) a. Write the equilibrium-constant expression for the reaction. b. Can you come up with some possible concentrations of \(\mathrm{C}\), \(\mathrm{D}\), and \(\mathrm{E}\) that you might observe when the reaction has reached equilibrium at \(25^{\circ} \mathrm{C} ?\) What are these values? c. A student says that only a very limited number of concentrations for \(\mathrm{C}, \mathrm{D}\), and \(\mathrm{E}\) are possible at equilibrium. Is this true? State why you think this is true or is not true. d. If you start with \(1.0 M\) concentrations of both \(\mathrm{C}\) and \(\mathrm{D}\) and allow the reaction to come to equilibrium, would you expect the concentration of \(\mathrm{C}\) to have decreased to zero? If not, what would you expect for the concentration of \(\mathrm{C} ?\) (An approximate value is fine.)

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