If the rate constant of a reaction is 0.03 s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The reaction is a first-order reaction because the unit of the rate constant is $s^{-1}$.For a first-order reaction,the time taken is given by the

Vedclass · Accepted Answer

$69.3$

Vedclass · Answer

(A) The reaction is a first-order reaction because the unit of the rate constant is $s^{-1}$.For a first-order reaction,the time taken is given by the formula: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.Given: $k = 0.03 \ s^{-1}$,$[A]_0 = 7.2 \ mol \ L^{-1}$,$[A]_t = 0.9 \ mol \ L^{-1}$.$t = \frac{2.303}{0.03} \log \frac{7.2}{0.9} = \frac{2.303}{0.03} \log 8$.Since $\log 8 = \log 2^3 = 3 \log 2 = 3 	imes 0.301 = 0.903$.$t = \frac{2.303 	imes 0.903}{0.03} \approx 69.3 \ s$.

If the rate constant of a reaction is $0.03 \ s^{-1}$,how much time does it take for $7.2 \ mol \ L^{-1}$ concentration of the reactant to get reduced to $0.9 \ mol \ L^{-1}$ (in $s$)? (Given: $\log 2 = 0.301$)

Similar Questions

$T_{50}$ of a first-order reaction is $10 \ min$. Starting with $10 \ mol \ L^{-1}$,the rate after $20 \ min$ is:

$A$ first-order reaction takes $69.3 \ min$ to complete $50\%$ of the reaction. How much time (in $min$) will it take to complete $80\%$ of the reaction?

The ratio of the time taken to complete $60\%$ and $20\%$ of a first-order reaction $[A \rightarrow \text{product}]$ is : $(\log 2 = 0.3)$

The decay constant of a first-order reaction is $1.1 \times 10^{-9} \ s^{-1}$. Calculate the half-life of the reaction.

The rate constant for a first order reaction is $60 \ s^{-1}$. How much time will it take to reduce the initial concentration of the reactant to its $1/16^{th}$ value?

Vedclass Test Series

Exam Paper Generator

Online Exam Module