Consider the following reaction:\[ 2\text{CH}_2\text{Cl}_2(g) \rightleftharpoons \text{CH}_4(g) + \text{CCl}_4(g) \]If 0.350 moles of \(\text{CH}_2\text{Cl}_2(g)\), 0.485 moles of \(\text{CH}_4\), and 0.269 moles of \(\text{CCl}_4\) are at equilibrium in a 10.7 L container at 541 K, the value of the equilibrium constant, \(K_c\), is ________.

GIVEN: At equilibrium Moles of CH2Cl2 = 0.350 moles Moles of CH4 =0.488 moles Moles of CCl4 = 0.269 mol Volume of the container=10.7 L Temperature =541 K To Calculate: Equilibrium constantWe know that, Concentration = Moles / Volume So, At equilibrium Concentration of CH2Cl2 = 0.350 moles / 10.7 L = 0.0327 M Concentration of CH4 =0.488 moles / 10.7 L =0.0456 M Concentration of CCl4 = 0.269 mol / 10.7 L =0.0251 M The given reaction is: 2 CH2Cl2 (g)<----------> CH4 (g) +CCl4 (g) The expression for the equilibrium constant for the given reaction can be written as: Kc =CH4CCl4CH2Cl22Now, substitute the equilibrium concentrations of all the speciesKc =0.04560.0251[0.0327]2 =0.001144560.00106929=1.07 ANSWER The value of the equilibrium constant is 1.07

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Consider the following reaction: 2CH,Cl½(g)CH,(g) + CC1,(3) If 0.350 moles of CH,C2(g), 0.488 moles of CH4, and 0.269 moles of CCl, are at equilibrium in a 10.7 L container at 541 K, the value of the equilibrium constant, K, is

Consider the following reaction: 2CH,Cl½(g)CH,(g) + CC1,(3) If 0.350 moles of CH,C2(g), 0.488 moles of CH4, and 0.269 moles of CCl, are at equilibrium in a 10.7 L container at 541 K, the value of the equilibrium constant, K, is

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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