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Chapter 15: Problem 94

At a temperature of 700 \(\mathrm{K}\) , the forward and reverse rate constants for the reaction \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)\) are \(k_{f}=1.8 \times 10^{-3} \mathrm{M}^{-1} \mathrm{s}^{-1}\) and \(k_{r}=0.063 \mathrm{M}^{-1} \mathrm{s}^{-1}\) (a) What is the value of the equilibrium constant \(K_{c}\) at 700 \(\mathrm{K} ?\) (b) Is the forward reaction endothermic or exothermic if the rate constants for the same reaction have values of \(k_{f}=0.097 M^{-1} \mathrm{s}^{-1}\) and \(k_{r}=2.6 \mathrm{M}^{-1} \mathrm{s}^{-1}\) at 800 \(\mathrm{K} ?\)

Short Answer

Expert verified
a) The equilibrium constant, Kc at 700K is 0.029. b) The forward reaction is endothermic.

Step by step solution

01

Part (a): Finding the equilibrium constant Kc

To find the value of the equilibrium constant, Kc, at 700K, we use the relationship between Kc and the forward and reverse rate constants given by: \[ K_c = \frac{k_f}{k_r} \] Given, kf = 1.8 x 10^-3 M^-1 s^-1 and kr = 0.063 M^-1 s^-1. Substitute these values into the formula: \[ K_c = \frac{1.8 \times 10^{-3}}{0.063} \]
02

Part (a): Calculating Kc

Now, divide 1.8 x 10^-3 by 0.063 to find the value of the equilibrium constant Kc: \[ K_c = 0.029 \] Thus, the equilibrium constant Kc at 700K is 0.029.
03

Part (b): Determining if the forward reaction is endothermic or exothermic

To determine if the forward reaction is endothermic or exothermic, we need to examine the rate constants at two different temperatures. We know that: 1. If the ratio of kf to kr increases as the temperature increases, the forward reaction is endothermic, i.e., it absorbs heat. 2. If the ratio of kf to kr decreases as the temperature increases, the forward reaction is exothermic, i.e., it releases heat. The given rate constants at 800K are kf = 0.097 M^-1 s^-1 and kr = 2.6 M^-1 s^-1. Calculate the ratio of kf to kr at 800K: \[ \frac{k_{f}(800K)}{k_{r}(800K)} = \frac{0.097}{2.6} \]
04

Part (b): Calculating the ratio of rate constants at 800K

Divide 0.097 by 2.6 to find the ratio of kf to kr at 800K: \[ \frac{k_{f}(800K)}{k_{r}(800K)} = 0.037 \] Now, compare this ratio to the Kc value we found at 700K (Kc = 0.029). Since the ratio of kf to kr at 800K is greater than the Kc value at 700K, the forward reaction is endothermic - it absorbs heat. In conclusion: a) The equilibrium constant Kc at 700K is 0.029. b) The forward reaction is endothermic.

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Most popular questions from this chapter

For the equilibrium $$2 \operatorname{IBr}(g) \Longrightarrow \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C} .\) If 0.025 atm of IBr is placed in a 2.0 -L container, what is the partial pressure of all substances after equilibrium is reached?

The water-gas shift reaction \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) is used industrially to produce hydrogen. The reaction enthalpy is \(\Delta H^{\circ}=-41 \mathrm{kJ}\) . (a) To increase the equilibrium yield of hydrogen would you use high or low temperature? ( b) Could you increase the equilibrium yield of hydrogen by controlling the pressure of this reaction? If so would high or low pressure favor formation of \(\mathrm{H}_{2}(g) ?\)

A mixture of 1.374 \(\mathrm{g}\) of \(\mathrm{H}_{2}\) and 70.31 \(\mathrm{g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00 -L vessel at 700 \(\mathrm{K}\) . These substances react according to $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ At equilibrium, the vessel is found to contain 0.566 \(\mathrm{g}\) of \(\mathrm{H}_{2}\) (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{Br}_{2},\) and \(\mathrm{HBr}\) . (b ) Calculate \(K_{c} .\)

At \(900^{\circ} \mathrm{C}, K_{\mathrm{C}}=0.0108\) 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? \begin{equation} \begin{array}{l}{\text { (a) } 15.0 \mathrm{g} \mathrm{CaCO}_{3}, 15.0 \mathrm{g} \mathrm{CaO}, \text { and } 4.25 \mathrm{gCO}_{2}} \\ {\text { (b) } 2.50 \mathrm{g} \mathrm{CaCO}_{3}, 25.0 \mathrm{g} \mathrm{CaO}, \text { and } 5.66 \mathrm{g} \mathrm{CO}_{2}} \\ {\text { (a) } 30.5 \mathrm{g} \mathrm{CaCO}_{3}, 25.5 \mathrm{g} \mathrm{CaO}, \text { and } 6.48 \mathrm{g} \mathrm{CO}_{2}}\end{array} \end{equation}

Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?
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