Question

For first order decomposition,

$A \longrightarrow 2B + 3C$

Concentration of A decreases from 1 M to 0.7 M in 10 seconds. What would be the concentration of A left after next 20 seconds.

• A. 0.1 M
• B. 0.147 M
• C. 0.21 M
• D. 0.343

Answer 1

Answer: (D)

0.343 M

Answer 2

For a first-order reaction, the concentration of a reactant decreases by a constant factor for every equal time interval. The integrated rate law for a first-order reaction is $[A]_t = [A]_0 e^{-kt}$.

Given: Initial concentration of A, $[A]_0 = 1$ M. Concentration of A after 10 seconds, $[A]_{10} = 0.7$ M.

1. Determine the decay factor for 10 seconds: The concentration decreases from 1 M to 0.7 M in 10 seconds. The factor by which concentration decreases is $\frac{[A]_{10}}{[A]_0} = \frac{0.7}{1} = 0.7$. This means $e^{-k \times 10 \text{ s}} = 0.7$.

2. Calculate the concentration after the total time: We need to find the concentration of A left after next 20 seconds. This means the total time elapsed from the beginning is $10 \text{ s} + 20 \text{ s} = 30 \text{ s}$. Let $[A]_{30}$ be the concentration of A after 30 seconds. Using the first-order integrated rate law: $[A]_{30} = [A]_0 e^{-k \times 30 \text{ s}}$ We can rewrite $e^{-k \times 30 \text{ s}}$ as $(e^{-k \times 10 \text{ s}})^3$. Substitute the decay factor for 10 seconds: $[A]_{30} = [A]_0 \times (0.7)^3$ $[A]_{30} = 1 \text{ M} \times (0.7 \times 0.7 \times 0.7)$ $[A]_{30} = 1 \text{ M} \times (0.49 \times 0.7)$ $[A]_{30} = 1 \text{ M} \times 0.343$ $[A]_{30} = 0.343$ M

The concentration of A left after next 20 seconds (total 30 seconds) is 0.343 M.