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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.

  

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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.

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Given, 

k = 2.0 × 10-2 s-1

T = 100 s

[A]o = 1.0 mol-1

Since the unit of k is s-1, the given reaction is a first order reaction.

Therefore, k = \(\frac{2.303}{t}\)log \(\frac{[A]_0}{[A]}\)

⇒ 2.0 x 10-2 s-1 = \(\frac{2.303}{100s}\)log\(\frac{1.0}{[A]}\)

⇒ 2.0 x 10-2 s-1 = \(\frac{2.303}{100s}\)log (-log[A])

⇒ -log[A] = \(\frac{2.0 \times 10^{-2} \times 100}{2.303}\)

⇒ [A] = antilog \(\Big(-\frac{2.0 \times 10^{-2} \times 100}{2.303}\Big)\)

= 0.135 mol L-1 (approximately)

Hence, the remaining concentration of A is 0.135 mol L-1.

This post was modified 6 months ago by admin
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