Question

The variation of concentration of 'A' with time in two experiments starting with two different initial concentration of 'A' is given by the following graph. The reaction is represented by A (aq) $\to$ B (aq). What is rate of reaction (M/min) when conc. of A in aqueous solution was 1.8 M?

• A. $0.072Mmi{n}^{-1}$
• B. $0.1296Mmi{n}^{-1}$
• C. $0.036Mmi{n}^{-1}$
• D. $1Mmi{n}^{-1}$

Answer 1

The correct option is A $0.072Mmi{n}^{-1}$

During first five minutes, the concentration decreases from 1.5 M to 1.2 M.
Thus, the rate of the reaction is $r_1=\frac{0.3}{5}Mmi{n}^{-1}$
During next 5 minutes, the concentration decreases from 1.2 M to 1.0 M
The rate of the reaction is $r_2=\frac{0.2}{5}$.
But $\frac{r_1}{r_2}={\left[\frac{A_1}{A_2}\right]}^m$
$\frac{0.3}{0.2}={\left(\frac{1.5}{1}\right)}^m$
$m=1$
The reaction is of the first order
The rate law expression is $\frac{0.3}{5}=k\left(1.5\right)$
The rate constant is $k=\frac{1}{25}$
Hence, the rate when the concentration is 1.8 M is $r=\frac{1}{25}\times 1.8=0.072Mmi{n}^{-1}$