The variation of concentration of A with time in two experiments starting with two different initial concentration of A is given in the following graph. The reaction is represented as A(aq) → B(aq). W

Question

The variation of concentration of A with time in two experiments starting with two different initial concentration of A is given in the following graph. The reaction is represented as A(aq) → B(aq). What is the rate of reaction $(M / min)$ when concentration of A in aqueous solution was $1.8 M ? (ln 5 = 1.609, ln 2 = 0.693)$

TestHub · Accepted Answer

Identify reaction order: Both experiments show the same concentration ratio from $t = 0$ to $t = 5 min$ $(\frac{1.5}{1.2} = \frac{1.0}{0.8} = 1.25)$ , confirming a first-order reaction.

Find rate constant ( $k$ ):

$k = \frac{1}{t} ln (\frac{[ A ] _{0}}{[ A ] _{t}}) = \frac{ln ( 1.25 )}{5} = \frac{ln 5 - 2 ln 2}{5} = \frac{1.609 - 1.386}{5} = 0.0446 mi n^{- 1}$

Calculate rate:

$R a t e = k [A] = 0.0446 \times 1.8 M \approx 0.08 Mmin^{- 1}$

Correct Answer: (C)

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