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Question: A first order reaction was commenced with 0.2M solution of the reactants. If the molarity of the sol...

A first order reaction was commenced with 0.2M solution of the reactants. If the molarity of the solution falls to 0.02M after 100 minutes the rate constant of the reaction is

A

2 x 10−2^{-2}min−1^{-1}

B

2.3 × 10−2^{-2}min−1^{-1}

C

4.6 × 10−2^{-2}min−1^{-1}

D

23 × 10−1^{-1}min−1^{-1}

Answer

2.3 × 10−2^{-2}min−1^{-1}

Explanation

Solution

The integrated rate law for a first-order reaction is given by:

k=2.303tlog⁡10[A]0[A]tk = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}

where kk is the rate constant, tt is the time, [A]0[A]_0 is the initial concentration of the reactant, and [A]t[A]_t is the concentration of the reactant at time tt.

Given values:

  • Initial concentration, [A]0=0.2[A]_0 = 0.2 M
  • Concentration after time tt, [A]t=0.02[A]_t = 0.02 M
  • Time, t=100t = 100 minutes

Substitute these values into the integrated rate law equation:

k=2.303100log⁡100.20.02k = \frac{2.303}{100} \log_{10} \frac{0.2}{0.02}

k=2.303100log⁡100.2×1000.02×100k = \frac{2.303}{100} \log_{10} \frac{0.2 \times 100}{0.02 \times 100}

k=2.303100log⁡10202k = \frac{2.303}{100} \log_{10} \frac{20}{2}

k=2.303100log⁡1010k = \frac{2.303}{100} \log_{10} 10

Since log⁡1010=1\log_{10} 10 = 1, the equation simplifies to:

k=2.303100×1k = \frac{2.303}{100} \times 1

k=2.303100k = \frac{2.303}{100}

k=0.02303 min−1k = 0.02303 \text{ min}^{-1}

k=2.303×10−2 min−1k = 2.303 \times 10^{-2} \text{ min}^{-1}