Question 2 of 69beginner🔧 ApplyNumerical2 marks

What is KcK_c for the following equilibrium when the equilibrium concentration of each substance is: [SO2]=0.60M,[O2]=0.82M[SO_2] = 0.60M, [O_2] = 0.82M and [SO3]=1.90M[SO_3] = 1.90M?

2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)
Correct Answer

Kc=12.23 M−1K_c = 12.23 \text{ M}^{-1}

Exercise: EXERCISES | Q: 6.2 | (Page 41)
For More Understanding

Explanation

Using the Law of Chemical Equilibrium, the equilibrium constant KcK_c is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients.

Solution Steps

  1. Step 1: Write the expression for KcK_c for the given reaction 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g).

    Kc=[SO3]2[SO2]2[O2]K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}

  2. Step 2: Substitute the given equilibrium concentrations into the expression. [SO2]=0.60 M[SO_2] = 0.60 \text{ M}, [O2]=0.82 M[O_2] = 0.82 \text{ M}, [SO3]=1.90 M[SO_3] = 1.90 \text{ M}

    Kc=(1.90)2(0.60)2×0.82K_c = \frac{(1.90)^2}{(0.60)^2 \times 0.82}

  3. Step 3: Calculate the value of KcK_c.

    Kc=3.610.36×0.82=3.610.2952≈12.23K_c = \frac{3.61}{0.36 \times 0.82} = \frac{3.61}{0.2952} \approx 12.23

  4. Step 4: Determine the units of KcK_c. Since the exponents of the numerator and denominator are not the same, KcK_c has a unit. Unit = (mol/L)2(mol/L)2(mol/L)=L mol−1\frac{(\text{mol/L})^2}{(\text{mol/L})^2 (\text{mol/L})} = \text{L mol}^{-1} or M−1\text{M}^{-1}

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