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Chapter 17: Problem 74

Predict the effect of increasing the temperature on the amount(s) of product(s) in the following reactions: (a) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\) \(\Delta H_{\mathrm{rxn}}^{\circ}=-90.7 \mathrm{~kJ}\) (b) \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=131 \mathrm{~kJ}\) (c) \(2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\) (endothermic) (d) \(2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) (exothermic)

Short Answer

Expert verified
(a) Decrease \(CH_3OH(g)\); (b) Increase \(CO(g)\) and \(H_2(g)\); (c) Increase \(NO(g)\) and \(O_2(g)\); (d) Decrease \(CO(g)\).

Step by step solution

01

Determine Reaction Type for (a)

Reaction (a) has \(\Delta H_{\mathrm{rxn}}^{\circ} = -90.7 \mathrm{~kJ}\), indicating it is exothermic. According to Le Chatelier's principle, increasing temperature shifts the equilibrium to the left (reactants).
02

Predict Product Change for (a)

Since (a) is exothermic, increasing the temperature decreases the amount of \(CH_3OH(g)\).
03

Determine Reaction Type for (b)

Reaction (b) has \(\Delta H_{\mathrm{rxn}}^{\circ} = 131 \mathrm{~kJ}\), indicating it is endothermic. Increasing temperature shifts the equilibrium to the right (products).
04

Predict Product Change for (b)

Since (b) is endothermic, increasing the temperature increases the amount of \(CO(g)\) and \(H_2(g)\).
05

Determine Reaction Type for (c)

Reaction (c) is stated to be endothermic. Increasing temperature shifts the equilibrium to the right (products).
06

Predict Product Change for (c)

Since (c) is endothermic, increasing the temperature increases the amount of \(NO(g)\) and \(O_2(g)\).
07

Determine Reaction Type for (d)

Reaction (d) is exothermic. Increasing temperature shifts the equilibrium to the left (reactants).
08

Predict Product Change for (d)

Since (d) is exothermic, increasing the temperature decreases the amount of \(CO(g)\).

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Key Concepts

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

exothermic reactions
Exothermic reactions release energy into their surroundings as heat or light. In these reactions, the reactants have more energy than the products. This means energy is given off during the reaction. We can identify an exothermic reaction in a chemical equation by a negative \(\Delta H\) value.

For example, an exothermic reaction looks like this:
\(\text{A} + \text{B} \rightarrow \text{C} + \text{D} \) with \(\Delta H = -90.7 \text{~kJ}\). Here, 'A' and 'B' release 90.7 kJ of energy when they form 'C' and 'D'.

According to Le Chatelier's Principle, if we increase the temperature of an exothermic reaction, the equilibrium position shifts to oppose this change. Since exothermic reactions release heat, adding more heat will shift the equilibrium towards the reactants. The system tries to absorb some of the extra heat by favoring the reverse reaction.
endothermic reactions
Endothermic reactions absorb energy from their surroundings. Unlike exothermic reactions, the products have more energy than the reactants. This means energy is taken in during the reaction. We can identify an endothermic reaction by a positive \(\Delta H\) value.

An example of an endothermic reaction looks like this:
\(\text{A} + \text{B} \rightarrow \text{C} + \text{D} \) with \(\Delta H = 131 \text{~kJ}\) Here, 'A' and 'B' absorb 131 kJ of energy to form 'C' and 'D'.

According to Le Chatelier's Principle, increasing the temperature of an endothermic reaction shifts the equilibrium position towards the products. Since endothermic reactions absorb heat, adding more heat will favor the forward reaction, helping the system take in the excess energy.
equilibrium shift
Le Chatelier's Principle explains how a system at equilibrium responds to changes. When a change occurs, the system shifts to counteract that change, seeking to establish a new equilibrium.

In the case of temperature changes:
  • For exothermic reactions, increasing the temperature shifts the equilibrium towards the reactants (left).
  • For endothermic reactions, increasing the temperature shifts the equilibrium towards the products (right).
This principle helps us understand how an equilibrium system will react to different conditions and can be applied to predict the effects on the position of equilibrium when conditions are changed.

Let's review the provided example reactions:
  • For reaction (a): Exothermic reaction shifting left on temperature increase.
  • For reaction (b): Endothermic reaction shifting right on temperature increase.
  • For reaction (c): Endothermic reaction shifting right on temperature increase.
  • For reaction (d): Exothermic reaction shifting left on temperature increase.

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Most popular questions from this chapter

As an EPA scientist studying catalytic converters and urban smog, you want to find \(K_{c}\) for the following reaction: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \quad K_{\mathrm{c}}=? $$ Use the following data to find the unknown \(K_{\mathrm{c}}\) : $$ \begin{aligned} \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \rightleftharpoons \mathrm{NO}(g) & & K_{\mathrm{c}}=4.8 \times 10^{-10} \\ 2 \mathrm{NO}_{2}(g) & \Longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & & K_{\mathrm{c}}=1.1 \times 10^{-5} \end{aligned} $$

The formation of methanol is important to the processing of new fuels. At \(298 \mathrm{~K}, K_{\mathrm{p}}=2.25 \times 10^{4}\) for the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(l) $$ If \(\Delta H_{\mathrm{rxn}}^{\circ}=-128 \mathrm{~kJ} / \mathrm{mol} \mathrm{CH}_{3} \mathrm{OH},\) calculate \(K_{\mathrm{p}}\) at \(0^{\circ} \mathrm{C}\)

Isopentyl alcohol reacts with pure acetic acid to form isopentyl acetate, the essence of banana oil: $$ \mathrm{C}_{5} \mathrm{H}_{11} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{5} \mathrm{H}_{11}+\mathrm{H}_{2} \mathrm{O} $$ A student adds a drying agent to remove \(\mathrm{H}_{2} \mathrm{O}\) and thus increase the yield of banana oil. Is this approach reasonable? Explain.

Isolation of Group \(8 \mathrm{~B}(10)\) elements, used as industrial catalysts, involves a series of steps. For nickel, the sulfide ore is roasted in air: \(\mathrm{Ni}_{3} \mathrm{~S}_{2}(s)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NiO}(s)+\mathrm{SO}_{2}(g) .\) The metal oxide is reduced by the \(\mathrm{H}_{2}\) in water gas \(\left(\mathrm{CO}+\mathrm{H}_{2}\right)\) to impure \(\mathrm{Ni}: \mathrm{NiO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{H}_{2} \mathrm{O}(g) .\) The \(\mathrm{CO}\) in water gas then reacts with the metal in the Mond process to form gaseous nickel carbonyl, \(\mathrm{Ni}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g),\) which is sub- sequently decomposed to the metal. (a) Balance each of the three steps, and obtain an overall balanced equation for the conversion of \(\mathrm{Ni}_{3} \mathrm{~S}_{2}\) to \(\mathrm{Ni}(\mathrm{CO})_{4}\). (b) Show that the overall \(Q_{\mathrm{c}}\) is the product of the \(Q_{c}\) 's for the individual reactions.

A change in reaction conditions increases the rate of a certain forward reaction more than that of the reverse reaction. What is the effect on the equilibrium constant and on the concentrations of reactants and products at equilibrium?
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