Question

A slow moving first order reaction is allowed to proceed for a long time. After $51$ minutes, one eighth of the original amount of reactant $A$ remains in the reaction vessel.
What is the rate constant of the reaction?

• A. $17$ minutes
• B. $0.0408$ minute
• C. $1.18$ minute
• D. $245$ minute
• E. $0.0136$ minute

Answer 1

The correct option is B $0.0408$ minute

For first order reaction,

$K=\frac{2.303}{t}\log \frac{\left[A_o\right]}{\left[A\right]}$

But $\left[A\right]=\frac{\left[A_o\right]}{8}\log \frac{\left[A_o\right]}{\left[A_o\right]}\times 8K=0.0408{min}^{-1}$