For a first order reaction,the time required | vedclass

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

Question

(C) For a first order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.The time $t$ required for a reaction to complete $90 \%$

Vedclass · Accepted Answer

$3.32$

Vedclass · Answer

(C) For a first order reaction,the rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}}$.The time $t$ required for a reaction to complete $90 \%$ is given by $t_{90\%} = \frac{2.303}{K} \log \left( \frac{100}{100 - 90} ight) = \frac{2.303}{K} \log 10 = \frac{2.303}{K}$.Substituting $K = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{t_{1/2}}$ into the equation:$t_{90\%} = \frac{2.303}{0.693} 	imes t_{1/2}$.Since $2.303 \approx \ln 10$ and $0.693 \approx \ln 2$,we have $t_{90\%} = \frac{\ln 10}{\ln 2} 	imes t_{1/2} = \frac{2.303}{2.303 	imes 0.3010} 	imes t_{1/2} = \frac{1}{0.3010} 	imes t_{1/2} \approx 3.32 	imes t_{1/2}$.Therefore,$x = 3.32$.

For a first order reaction,the time required for completion of $90 \%$ reaction is '$x$' times the half life of the reaction. The value of '$x$' is $........$. (Given: $\ln 10 = 2.303$ and $\log 2 = 0.3010$)

Similar Questions

Decomposition of $H_2O_2$ follows a first order reaction. In $50 \ min$,the concentration of $H_2O_2$ decreases from $0.5 \ M$ to $0.125 \ M$. For such decomposition,when the concentration of $H_2O_2$ reaches $0.05 \ M$,the rate of formation of $O_2$ will be:

The velocity constant of a first-order reaction is expressed in the units:

The reaction $2H_2O_2 \to 2H_2O + O_2$ is a

If the rate constant for a first order reaction is $k$,the time $(t)$ required for the completion of $99 \%$ of the reaction is given by

Calculate the half-life of a first-order reaction in minutes if the rate constant is $1 \times 10^{-3} \ sec^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module