Question 33 of 69beginner🔧 ApplyNumerical2 marks

The value of KcK_c for the reaction 3O2(g)⇌2O3(g)3O_2(g) \rightleftharpoons 2O_3(g) is 2.0×10−502.0 \times 10^{-50} at 25°C. If the equilibrium concentration of O2_2 in air at 25°C is 1.6×10−21.6 \times 10^{-2}, what is the concentration of O3_3?

Correct Answer

2.86×10−282.86 \times 10^{-28} M

Exercise: EXERCISES | Q: 6.33 | (Page 45)
For More Understanding

Explanation

To find the concentration of O3_3, we use the equilibrium constant expression for the given reaction. By substituting the known values of KcK_c and the equilibrium concentration of O2_2, we can algebraically solve for the unknown concentration of O3_3.

Solution Steps

  1. Step 1: Write the expression for the equilibrium constant (KcK_c) for the reaction 3O2(g)⇌2O3(g)3O_2(g) \rightleftharpoons 2O_3(g).

    Kc=[O3]2[O2]3K_c = \frac{[O_3]^2}{[O_2]^3}

  2. Step 2: Substitute the given values into the expression. We are given Kc=2.0×10−50K_c = 2.0 \times 10^{-50} and [O2]=1.6×10−2[O_2] = 1.6 \times 10^{-2}.

    2.0×10−50=[O3]2(1.6×10−2)32.0 \times 10^{-50} = \frac{[O_3]^2}{(1.6 \times 10^{-2})^3}

  3. Step 3: Calculate the value of [O2]3[O_2]^3.

    (1.6×10−2)3=4.096×10−6(1.6 \times 10^{-2})^3 = 4.096 \times 10^{-6}

  4. Step 4: Rearrange the equation to solve for [O3]2[O_3]^2.

    [O3]2=(2.0×10−50)×(4.096×10−6)=8.192×10−56[O_3]^2 = (2.0 \times 10^{-50}) \times (4.096 \times 10^{-6}) = 8.192 \times 10^{-56}

  5. Step 5: Take the square root of both sides to find the concentration of O3O_3. [O3]=8.192×10−56=2.86×10−28[O_3] = \sqrt{8.192 \times 10^{-56}} = 2.86 \times 10^{-28} M

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