The equilibrium constant, K₉, for the following reaction is 9.52×10⁻² at 350 K.\[ \text{CH}_4 \, (\text{g}) + \text{CCl}_4 \, (\text{g}) \rightleftharpoons 2 \, \text{CH}_2\text{Cl}_2 \, (\text{g}) \]Calculate the equilibrium concentrations of reactants and product when 0.289 moles of CH₄ and 0.289 moles of CCl₄ are introduced into a 1.00 L vessel at 350 K.\[ [\text{CH}_4] = \boxed{\hspace{1cm}} \, \text{M} \]\[ [\text{CCl}_4] = \boxed{\hspace{1cm}} \, \text{M} \]\[ [\text{CH}_2\text{Cl}_2] = \boxed{\hspace{1cm}} \, \text{M} \] ### Equilibrium Concentration CalculationThe equilibrium constant, \( K_c \), for the following reaction is **10.5** at **350 K**.#### Reaction:\[ 2 \text{CH}_2\text{Cl}_2 (g) \rightleftharpoons \text{CH}_4 (g) + \text{CCl}_4 (g) \]#### Problem:Calculate the equilibrium concentrations of reactant and products when **0.328 moles** of \(\text{CH}_2\text{Cl}_2\) are introduced into a **1.00 L** vessel at **350 K**.#### Equilibrium Concentrations:- \([\text{CH}_2\text{Cl}_2]\) = \([ \, ] \, \text{M}\)- \([\text{CH}_4]\) = \([ \, ] \, \text{M}\)- \([\text{CCl}_4]\) = \([ \, ] \, \text{M}\)

The equilibrium constant, K., for the following reaction is 9.52×10-2 at 350 K. CH4 (g) + CC14 (g)=2 CH2C12 (g) Calculate the equilibrium concentrations of reactants and product when 0.289 moles of CH4 and [ CH4] [ CC4] M M

The equilibrium constant, K., for the following reaction is 9.52×10-2 at 350 K. CH4 (g) + CC14 (g)=2 CH2C12 (g) Calculate the equilibrium concentrations of reactants and product when 0.289 moles of CH4 and [ CH4] [ CC4] M M

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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