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Chemical Kinetics

Question

Consider a certain reaction A → Products with K = 2.0 x 10^–2 s^–1. Calculate the concentration of A remaining after 100 S if the initial concentration of A is 1.0 mol L^–1.

Solution

k= 2.0 x10^-2s^-1
T=100
[A]₀ =1.0 mol^-1
The givaen reaction is first order reaction

K = 2.0 \times 10^{- 2} s^{- 1} t_{1 / 2} = 100 \sec, a = 1.0 mol L^{- 1} K = \frac{2.303}{t} \log \frac{a}{a - x} 2.0 \times 10^{- 2} = \frac{2.303}{100} \log \frac{1.0}{1.0 - x} or \frac{2.0 \times 10^{- 2} \times 100}{2.303} = \log \frac{1}{1 - x} or 0.8684 = \log \frac{1}{1 - x} Antilog 0.8684 = \log \frac{1}{1 - x} or 7.386 = \frac{1}{1 - x} or 7.386 (1 - x) = 1 or 1 - x = \frac{1}{7.386} = 0.135

Hence, concentration of A remaining after 100 S = 0.135 M.