### Chemical Equilibrium and Temperature DependenceThis 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- \( \Delta H^\circ = 16.1 \times 10^3 \) JFirst, 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

Given that: K1 = 0.160 T1 = 298 K T2 = 420 K ΔH = 16.1 kJ= 16100 J To find the values of…

Answered: The equilibrium constant, K,, for the…

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 =

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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