The equilibrium constant, K,, for the following reaction is 0.160 at 298 K. 2NOBr(g) 2NO(g) + Br2(g) If AH° for this reaction is 16.1 kJ, what is the value of K, at 420 K? Kp =
The equilibrium constant, K,, for the following reaction is 0.160 at 298 K. 2NOBr(g) 2NO(g) + Br2(g) If AH° for this reaction is 16.1 kJ, what is the value of K, at 420 K? Kp =
Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter7: Reaction Rates And Chemical Equilibrium
Section: Chapter Questions
Problem 7.81P
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![### Chemical Equilibrium and Temperature Dependence
This section of the educational material discusses the equilibrium constant and its temperature dependence for the following chemical reaction:
\[ \text{2NOBr(g)} \rightleftharpoons \text{2NO(g) + Br}_2\text{(g)} \]
#### Given Data:
- The equilibrium constant (\( K_p \)) at 298 K is 0.160.
- The enthalpy change (\( \Delta H^\circ \)) for this reaction is 16.1 kJ.
- We are required to find the equilibrium constant (\( K_p \)) at 420 K.
#### Detailed Explanation:
To determine the value of \( K_p \) at 420 K, the Van't Hoff equation can be used:
\[ \ln \left( \dfrac{K_2}{K_1} \right) = -\dfrac{\Delta H^\circ}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \]
Where:
- \( K_1 \) and \( K_2 \) are the equilibrium constants at temperatures \( T_1 \) and \( T_2 \), respectively.
- \( R \) is the universal gas constant (8.314 J/(mol·K)).
- \( \Delta H^\circ \) is the enthalpy change.
Rearranging the equation to isolate \( K_2 \):
\[ K_2 = K_1 \exp \left( -\dfrac{\Delta H^\circ}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \right) \]
Using the provided data:
- \( K_1 = 0.160 \)
- \( T_1 = 298 \) K
- \( T_2 = 420 \) K
- \( \Delta H^\circ = 16.1 \times 10^3 \) J
First, calculate the term \( \dfrac{1}{T_2} - \dfrac{1}{T_1} \):
\[ \dfrac{1}{420} - \dfrac{1}{298} = -0.002679 \text{ K}^{-1} \]
Then, calculate \( \dfrac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a2a9461-f2b7-4bf6-ae50-440611acf6b0%2F974b6ab9-2703-4139-a058-375bd0ca0082%2Fkkykutzi_processed.png&w=3840&q=75)
Transcribed Image Text:### Chemical Equilibrium and Temperature Dependence
This section of the educational material discusses the equilibrium constant and its temperature dependence for the following chemical reaction:
\[ \text{2NOBr(g)} \rightleftharpoons \text{2NO(g) + Br}_2\text{(g)} \]
#### Given Data:
- The equilibrium constant (\( K_p \)) at 298 K is 0.160.
- The enthalpy change (\( \Delta H^\circ \)) for this reaction is 16.1 kJ.
- We are required to find the equilibrium constant (\( K_p \)) at 420 K.
#### Detailed Explanation:
To determine the value of \( K_p \) at 420 K, the Van't Hoff equation can be used:
\[ \ln \left( \dfrac{K_2}{K_1} \right) = -\dfrac{\Delta H^\circ}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \]
Where:
- \( K_1 \) and \( K_2 \) are the equilibrium constants at temperatures \( T_1 \) and \( T_2 \), respectively.
- \( R \) is the universal gas constant (8.314 J/(mol·K)).
- \( \Delta H^\circ \) is the enthalpy change.
Rearranging the equation to isolate \( K_2 \):
\[ K_2 = K_1 \exp \left( -\dfrac{\Delta H^\circ}{R} \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \right) \]
Using the provided data:
- \( K_1 = 0.160 \)
- \( T_1 = 298 \) K
- \( T_2 = 420 \) K
- \( \Delta H^\circ = 16.1 \times 10^3 \) J
First, calculate the term \( \dfrac{1}{T_2} - \dfrac{1}{T_1} \):
\[ \dfrac{1}{420} - \dfrac{1}{298} = -0.002679 \text{ K}^{-1} \]
Then, calculate \( \dfrac
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