**Reaction Quotient and Equilibrium Constant****Consider this equilibrium reaction at 400 K:**\[ \text{Br}_2(g) + \text{Cl}_2(g) \rightleftharpoons 2 \text{BrCl}(g) \quad K_c = 7.0 \]If the composition of the reaction mixture at 400 K is $[\text{BrCl}] = 0.00248 \, \text{M}$, $[\text{Br}_2] = 0.00570 \, \text{M}$, and $[\text{Cl}_2] = 0.00105 \, \text{M}$, what is the reaction quotient, $Q$?\[ Q = \]**How is the reaction quotient related to the equilibrium constant, $K_c$, for this reaction?**- $Q < K_c$- $Q > K_c$- $Q = K_c$**Explanation of Reaction Quotient, $Q$:**The reaction quotient, $Q$, provides a snapshot of the ratio of the concentrations of products to reactants at any point in time for a reversible reaction. It is calculated in a similar way to the equilibrium constant, $K_c$, but the concentrations do not have to be at equilibrium.For the given reaction:\[ \text{Br}_2(g) + \text{Cl}_2(g) \rightleftharpoons 2 \text{BrCl}(g) \]The reaction quotient, $Q$, is given by:\[ Q = \frac{[\text{BrCl}]^2}{[\text{Br}_2][\text{Cl}_2]} \]Given:- $[\text{BrCl}] = 0.00248 \, \text{M}$- $[\text{Br}_2] = 0.00570 \, \text{M}$- $[\text{Cl}_2] = 0.00105 \, \text{M}$Substituting these values into the equation for $Q$:\[ Q = \frac{(0.00248)^2}{(0.00570)(0.00105)} \]Calculate $Q$ and compare it to $K_c$ to determine if the system is at equilibrium ($Q = K_c$),

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Consider this equilibrium reaction at 400 K. Br, (g) + Cl,(g) = 2 BrCI(g) K. = 7.0 If the composition of the reaction mixture at 400 K is [BrCI] = 0,00248 M, [Br,] = 0.00570 M, and [Cl,] = 0.00105 M, what is the reaction quotient, Q? Q = How is the reaction quotient related to the equilibrium constant, Ke, for this reaction? Oo < K. O e > K, O0 = K.

Consider this equilibrium reaction at 400 K. Br, (g) + Cl,(g) = 2 BrCI(g) K. = 7.0 If the composition of the reaction mixture at 400 K is [BrCI] = 0,00248 M, [Br,] = 0.00570 M, and [Cl,] = 0.00105 M, what is the reaction quotient, Q? Q = How is the reaction quotient related to the equilibrium constant, Ke, for this reaction? Oo < K. O e > K, O0 = K.

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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