Calculate the value of K, in the following reaction if the equilibrium concentrations are [H,S] = 0.25 M, [H,] = 0.88 M, [S,] = 0.44M. 2 H,Se = 2 Hzie) '2(g)

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### Calculating the Equilibrium Constant (K_eq)

To determine the value of the equilibrium constant (\(K_{eq}\)) for the given reaction, we use the equilibrium concentrations provided.

The reaction in question is:
\[ 2 \text{H}_2\text{S}_{(g)} \rightleftharpoons 2 \text{H}_{2(g)} + \text{S}_{2(g)} \]

The equilibrium concentrations are:
- \([\text{H}_2\text{S}] = 0.25 \, \text{M}\)
- \([\text{H}_2] = 0.88 \, \text{M}\)
- \([\text{S}_2] = 0.44 \, \text{M}\)

### Steps to Calculate \(K_{eq}\)

1. **Write the expression for the equilibrium constant (\(K_{eq}\)):**

   \[
   K_{eq} = \frac{[\text{H}_2]^2 [\text{S}_2]}{[\text{H}_2\text{S}]^2}
   \]

2. **Substitute the equilibrium concentrations into the expression:**

   \[
   K_{eq} = \frac{(0.88)^2 \cdot (0.44)}{(0.25)^2}
   \]

3. **Perform the calculations:**
   
   - Calculate the numerator: \((0.88)^2 \times (0.44) = 0.7744 \times 0.44 = 0.340736\)
   - Calculate the denominator: \((0.25)^2 = 0.0625\)

   \[
   K_{eq} = \frac{0.340736}{0.0625} \approx 5.45
   \]

### Result

The value of the equilibrium constant (\(K_{eq}\)) for the reaction is approximately **5.45**.

### Explanation of Diagrams and Graphs

There are no diagrams or graphs included in this content. The text focuses on presenting and solving a chemical equilibrium problem through mathematical expressions and equilibrium concentrations.
Transcribed Image Text:### Calculating the Equilibrium Constant (K_eq) To determine the value of the equilibrium constant (\(K_{eq}\)) for the given reaction, we use the equilibrium concentrations provided. The reaction in question is: \[ 2 \text{H}_2\text{S}_{(g)} \rightleftharpoons 2 \text{H}_{2(g)} + \text{S}_{2(g)} \] The equilibrium concentrations are: - \([\text{H}_2\text{S}] = 0.25 \, \text{M}\) - \([\text{H}_2] = 0.88 \, \text{M}\) - \([\text{S}_2] = 0.44 \, \text{M}\) ### Steps to Calculate \(K_{eq}\) 1. **Write the expression for the equilibrium constant (\(K_{eq}\)):** \[ K_{eq} = \frac{[\text{H}_2]^2 [\text{S}_2]}{[\text{H}_2\text{S}]^2} \] 2. **Substitute the equilibrium concentrations into the expression:** \[ K_{eq} = \frac{(0.88)^2 \cdot (0.44)}{(0.25)^2} \] 3. **Perform the calculations:** - Calculate the numerator: \((0.88)^2 \times (0.44) = 0.7744 \times 0.44 = 0.340736\) - Calculate the denominator: \((0.25)^2 = 0.0625\) \[ K_{eq} = \frac{0.340736}{0.0625} \approx 5.45 \] ### Result The value of the equilibrium constant (\(K_{eq}\)) for the reaction is approximately **5.45**. ### Explanation of Diagrams and Graphs There are no diagrams or graphs included in this content. The text focuses on presenting and solving a chemical equilibrium problem through mathematical expressions and equilibrium concentrations.
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