Question

The rate constant for an isomerisation reaction, $A\to B$ is $4.5\times {10}^{-3}mi{n}^{-1}$. If the initial concentration of A is 1M, calculate the rate of reaction after 1 h.

• A. $3.44\times {10}^{-3}mol{L}^{-1}mi{n}^{-1}$
• B. $3.44\times {10}^{3}mol{L}^{-1}mi{n}^{-1}$
• C. $1.86\times {10}^{-3}mol{L}^{-1}mi{n}^{-1}$
• D. $1.86\times {10}^{3}mol{L}^{-1}mi{n}^{-1}$

Answer 1

The correct option is A $3.44\times {10}^{-3}mol{L}^{-1}mi{n}^{-1}$

Rate constant in min${}^{-1}$ indicates that, the reaction is of 1st order. Hence, $k=\frac{2.303}{t}log\frac{[R{]}_0}{\left[R\right]}$

$4.5\times {10}^{-3}mi{n}^{-1}=\frac{2.303}{60min}log\left(\frac{1}{\left[R\right]}\right)$

$log\left[R\right]=-0.1172=\overline{1}.8828$

or $\left[R\right]=$ antilong $\overline{1}.8828$

$\left[A\right]=0.7635mol{L}^{-1}$; this is the concentration after 60 min or one hour.

So, rate after 1 hour $Rate=K\left[A\right]=4.5\times {10}^{-3}\times 0.765$

=$3.44\times {10}^{-3}mol{L}^{-1}mi{n}^{-1}$