### Reaction Equilibrium Calculation at 586 KA student conducted an experiment to study the following reaction at a temperature of 586 K:\[ \text{CO(g)} + \text{Cl}_2\text{(g)} \rightleftharpoons \text{COCl}_2\text{(g)} \]#### Initial Conditions- **Moles of CO(g)**: 0.376 moles- **Moles of Cl₂(g)**: 0.406 moles- **Volume of container**: 1.00 liter#### Observed Equilibrium Condition- **Equilibrium concentration of COCl₂(g)**: 0.337 M#### Calculation of Equilibrium Constant \( K_c \)To find the equilibrium constant \( K_c \), we need to set up the reaction table and expressions for equilibrium concentrations and then apply the equilibrium constant formula.1. **Initial Concentrations (in Moles per Liter):** - \([ \text{CO} ]_0 = \frac{0.376 \text{ moles}}{1.00 \text{ L}} = 0.376 \text{ M}\) - \([ \text{Cl}_2 ]_0 = \frac{0.406 \text{ moles}}{1.00 \text{ L}} = 0.406 \text{ M}\)2. **Determination of Changes:** - At equilibrium, \[\text{COCl}_2 \text{(g)} = 0.337 \text{ M}\] - Assuming \(x\) moles of CO and \(x\) moles of Cl₂ react to form 0.337 M of the product, then: - \([ \text{CO} ] = 0.376\text{ M} - x\] - \([ \text{Cl}_2 ] = 0.406\text{ M} - x\] - \([ \text{COCl}_2 ] = 0.337 \text{ M}\) - Since \(x = 0.337 \text{ M}\), the remaining concentrations are: - \([ \text{CO} ] = 0.376\text{ M} - 0.337\text{ M} = 0.039 \text{ M}\)

Molarity = number of moles/ volume in Litre Molarity of CO = 0.376/1 = 0.376M Molarity of Cl2 =…

A student ran the following reaction in the laboratory at 586 K: CO(g) + Cl2(g) COCI,(g) When she introduced 0.376 moles of CO(g) and 0.406 moles of Cl½(g) into a 1.00 liter container, she found the equilibrium concentration of COCI,(g) to be 0.337 М. Calculate the equilibrium constant, K, she obtained for this reaction. K =

A student ran the following reaction in the laboratory at 586 K: CO(g) + Cl2(g) COCI,(g) When she introduced 0.376 moles of CO(g) and 0.406 moles of Cl½(g) into a 1.00 liter container, she found the equilibrium concentration of COCI,(g) to be 0.337 М. Calculate the equilibrium constant, K, she obtained for this reaction. 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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