The equilibrium constant, \( K_c \), for the following reaction is 0.0120 at 500 K.\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]If an equilibrium mixture of the three gases in an 18.1 L container at 500 K contains 0.224 mol of \( \text{PCl}_5(g) \) and 0.420 mol of \( \text{PCl}_3 \), the equilibrium concentration of \( \text{Cl}_2 \) is \(\boxed{\phantom{2}}\) M. **Equilibrium Reaction Analysis**Consider the following reaction:\[ \text{PCl}_5 (g) \rightleftharpoons \text{PCl}_3 (g) + \text{Cl}_2 (g) \]Given the conditions: - 0.113 moles of \(\text{PCl}_5 (g)\)- 0.640 moles of \(\text{PCl}_3 (g)\)- 0.430 moles of \(\text{Cl}_2 (g)\)These are at equilibrium in a 19.4 L container at a temperature of 566 K. The task is to calculate the equilibrium constant, \( K_c \), for the reaction under these conditions. **Calculations Needed:**1. **Concentrations:** Calculate the molarity of each component by dividing the moles by the volume of the container in liters. \[ [\text{PCl}_5] = \frac{0.113}{19.4} \, \text{mol/L} \] \[ [\text{PCl}_3] = \frac{0.640}{19.4} \, \text{mol/L} \] \[ [\text{Cl}_2] = \frac{0.430}{19.4} \, \text{mol/L} \]2. **Equilibrium Constant Expression:** The equilibrium constant \( K_c \) expression for the reaction is: \[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \] Substitute the concentrations calculated to find the value of \( K_c \). **Objective:** Determine the exact value of \( K_c \) using the calculated concentrations. This reflects the extent to which the reaction favors the products at a given temperature (566 K).

Answer:For any reaction, value of equilibrium constant KC is equal to the ratio of molar…

The equilibrium constant, Kc, for the following reaction is 0.0120 at 500 K. PC15 (9) PC13 (g) + Cl₂ (9) If an equilibrium mixture of the three gases in a 18.1 L container at 500 K contains 0.224 mol of PC15 (9) and 0.420 mol of PC13, the equilibrium concentration of Cl₂ is | M.

The equilibrium constant, Kc, for the following reaction is 0.0120 at 500 K. PC15 (9) PC13 (g) + Cl₂ (9) If an equilibrium mixture of the three gases in a 18.1 L container at 500 K contains 0.224 mol of PC15 (9) and 0.420 mol of PC13, the equilibrium concentration of Cl₂ is | 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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