**Equilibrium Constant Calculation for Bromine Gas Dissociation****Problem Statement:**Consider the following reaction at a high temperature:\[ \text{Br}_2(g) \rightleftharpoons 2\text{Br}(g) \]When 1.30 moles of \(\text{Br}_2\) are put in a 0.710-L flask, 1.90 percent of the \(\text{Br}_2\) undergoes dissociation. Calculate the equilibrium constant \(K_c\) for the reaction.**Calculation Steps:**1. **Initial Concentration:** \[ [\text{Br}_2]_{\text{initial}} = \frac{1.30 \text{ moles}}{0.710 \text{ L}} = 1.83 \text{ M} \]2. **Dissociation Details:** 1.90% of \(\text{Br}_2\) dissociates: \[ \text{Moles of } \text{Br}_2 \text{dissociated} = 1.30 \text{ moles} \times 0.019 = 0.0247 \text{ moles} \] \[ \text{Concentration of } \text{Br}_2 \text{dissociated} = \frac{0.0247 \text{ moles}}{0.710 \text{ L}} = 0.0348 \text{ M} \]3. **Change in Concentration:** For every mole of \(\text{Br}_2\) that dissociates, 2 moles of \(\text{Br}\) are produced: \[ [\text{Br}] = 2 \times 0.0348 \text{ M} = 0.0696 \text{ M} \] \[ [\text{Br}_2]_{\text{equilibrium}} = [\text{Br}_2]_{\text{initial}} - [\text{Br}_2 \text{dissociated}] = 1.83 \text{ M} - 0.0348 \text{ M} = 1.80 \text{ M} \]4. **Equilibrium

Given, Br2g ⇔ 2Brg Moles of Br2 put in flask = 1.30 moles Volume of flask = 0.710 L Percent…

Consider the following reaction at a high temperature. Br₂(g) 2Br(g) When 1.30 moles of Br2 are put in a 0.710-L flask, 1.90 percent of the Br2 Calculate the equilibrium constant Ke for the reaction.

Consider the following reaction at a high temperature. Br₂(g) 2Br(g) When 1.30 moles of Br2 are put in a 0.710-L flask, 1.90 percent of the Br2 Calculate the equilibrium constant Ke for the reaction.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter12: Chemical Equilibrium

Section: Chapter Questions

Problem 92QRT

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