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question 4 (1 point)
sulfur dioxide is a colourless gas, which smells like burnt matches. when sulfur dioxide and chlorine react with each other, they form so₂cl₂ as
so₂(g) + cl₂(g) ⇌ so₂cl₂(g)
at equilibrium, the concentration values of compounds are:
| compound | so₂(g) | cl₂(g) | so₂cl₂(g) |
|---|
the value of equilibrium constant k_eq, for the above reaction is
○ 0.72
○ 0.75
○ 0.80
○ 0.82
○ 1.33
Step1: Recall the formula for \( K_{eq} \)
For the reaction \( \text{SO}_2(\text{g}) + \text{Cl}_2(\text{g})
ightleftharpoons \text{SO}_2\text{Cl}_2(\text{g}) \), the equilibrium constant expression is \( K_{eq} = \frac{[\text{SO}_2\text{Cl}_2]}{[\text{SO}_2][\text{Cl}_2]} \).
Step2: Substitute the given concentrations
We know \( [\text{SO}_2] = 1.78 \, \text{mol/L} \), \( [\text{Cl}_2] = 0.90 \, \text{mol/L} \), and \( [\text{SO}_2\text{Cl}_2] = 1.20 \, \text{mol/L} \). Substituting these into the formula:
\( K_{eq} = \frac{1.20}{(1.78)(0.90)} \)
Step3: Calculate the denominator
First, calculate \( (1.78)(0.90) = 1.602 \)
Step4: Calculate \( K_{eq} \)
Now, divide 1.20 by 1.602: \( K_{eq} = \frac{1.20}{1.602} \approx 0.75 \)
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0.75 (corresponding to the option with 0.75)