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Chapter 15: Problem 70

True or false: When the temperature of an exothermic reaction increases, the rate constant of the forward reaction decreases, which leads to a decrease in the equilibrium constant, \(K_{c}\)

Short Answer

Expert verified
The given statement is partially true and partially false. When the temperature of an exothermic reaction increases, the rate constant of the forward reaction actually increases, not decreases, as given by the Arrhenius equation. However, it is true that an increase in temperature results in a decrease in the equilibrium constant, \(K_{c}\), for an exothermic reaction, as described by the van't Hoff equation.

Step by step solution

01

Understanding Exothermic Reactions

An exothermic reaction is a chemical reaction that releases energy in the form of heat. According to Le Chatelier's principle, increasing the temperature of an exothermic reaction will shift the equilibrium position to favor the reverse (endothermic) reaction. This is because the system will try to counteract the increased temperature by absorbing the extra heat.
02

Effect on Rate Constants

The rate constant, denoted by \(k\), is affected by temperature changes according to the Arrhenius equation: \[k = A \exp\left(\frac{-E_{a}}{RT}\right)\] where \(A\) is the pre-exponential factor, \(E_{a}\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. As the temperature increases, the exponential term in the equation will generally decrease, which will result in an increase in the rate constant. Hence, the statement "When the temperature of an exothermic reaction increases, the rate constant of the forward reaction decreases" is false.
03

Effect on the Equilibrium Constant

The equilibrium constant, \(K_{c}\), also depends on temperature, as given by the van't Hoff equation: \[\frac{d\ln K_{c}}{dT} = \frac{\Delta H_{rxn}}{RT^2}\] where \(\Delta H_{rxn}\) is the enthalpy change of the reaction. Since the reaction is exothermic, \(\Delta H_{rxn}\) is negative. Therefore, for an exothermic reaction, increasing the temperature will result in a decrease in the equilibrium constant. Thus, the statement "which leads to a decrease in the equilibrium constant, \(K_{c}\)" is true.
04

Conclusion

The initial statement "When the temperature of an exothermic reaction increases, the rate constant of the forward reaction decreases, which leads to a decrease in the equilibrium constant, \(K_{c}\)" is partially true and partially false. The rate constant of the forward reaction actually increases with increasing temperature. However, it is true that increasing the temperature of an exothermic reaction results in a decrease in the equilibrium constant, \(K_{c}\).

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Most popular questions from this chapter

If \(K_{c}=0.013 \mathrm{~L} / \mathrm{mol}\) for $2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$ at \(1000 \mathrm{~K}\), what is the value of \(K_{p}\) for this reaction at this temperature?

(a) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

A flask is charged with \(152.0 \mathrm{kPa}\) of $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( and \)101.3 \mathrm{kPa}\( \)\mathrm{NO}_{2}(g)$ at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is \(51.9 \mathrm{kPa}\). (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?(\mathbf{b})\) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

Suppose that the gas-phase reactions \(A \longrightarrow B\) and $B \longrightarrow A\( are both elementary processes with rate constants of \)2.5 \times 10^{-2} \mathrm{~min}^{-1}\( and \)2.5 \times 10^{-1} \mathrm{~min}^{-1}$, respectively. (a) What is the value of the equilibrium constant for the equilibrium $\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?(\mathbf{b})$ Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (a) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)
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