Suppose a 250. mL flask is filled with 1.2 mol of Cl₂ and 0.60 mol of HCl. The following reaction becomes possible: H₂(g) + Cl₂(g) 2HCl(g) The equilibrium constant K for this reaction is 0.409 at the temperature of the flask. Calculate the equilibrium molarity of H₂. Round your answer to two decimal places. M X S

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### Kinetics and Equilibrium: Calculating Equilibrium Composition from an Equilibrium Constant

**Problem Statement:**

Suppose a 250. mL flask is filled with 1.2 mol of Cl₂ and 0.60 mol of HCl. The following reaction becomes possible:

\[ \text{H}_2(g) + \text{Cl}_2(g) \rightleftharpoons 2\text{HCl}(g) \]

The equilibrium constant \( K \) for this reaction is 0.409 at the temperature of the flask.

**Task:**

Calculate the equilibrium molarity of \( \text{H}_2 \). Round your answer to two decimal places.

**Input Box:** [ \[ \text{M} \] ]

**Buttons:**
- Explanation
- Check

**Note:**
This problem involves calculating how the concentrations of reactants and products change until they reach equilibrium, using the provided equilibrium constant.
Transcribed Image Text:### Kinetics and Equilibrium: Calculating Equilibrium Composition from an Equilibrium Constant **Problem Statement:** Suppose a 250. mL flask is filled with 1.2 mol of Cl₂ and 0.60 mol of HCl. The following reaction becomes possible: \[ \text{H}_2(g) + \text{Cl}_2(g) \rightleftharpoons 2\text{HCl}(g) \] The equilibrium constant \( K \) for this reaction is 0.409 at the temperature of the flask. **Task:** Calculate the equilibrium molarity of \( \text{H}_2 \). Round your answer to two decimal places. **Input Box:** [ \[ \text{M} \] ] **Buttons:** - Explanation - Check **Note:** This problem involves calculating how the concentrations of reactants and products change until they reach equilibrium, using the provided equilibrium constant.
Expert Solution
Step 1

Given ->

H2(g) + Cl2(g) <-----> 2HCl(g) 

Mole of Cl2 = 1.2 mole 

Mole of HCl = 0.60 mole 

Equilibrium constant (Keq) = 0.409 

Volume= 250 ml = 0.250 L (1L = 1000ml) 

 

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