If benzene diazonium chloride undergoes first | 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 equation is given by:$t = \frac{2.303}{k} \log \frac{a}{a-x}$Given that the reaction is $90 \%$ complete,we ha

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(C) For a first order reaction,the rate equation is given by:$t = \frac{2.303}{k} \log \frac{a}{a-x}$Given that the reaction is $90 \%$ complete,we have $a = 100$ and $(a-x) = 100 - 90 = 10$.The rate constant $k = 6.93 	imes 10^{-2} \ min^{-1}$.Substituting these values into the equation:$t = \frac{2.303}{6.93 	imes 10^{-2}} \log \frac{100}{10}$$t = \frac{2.303}{0.0693} 	imes \log(10)$Since $\log(10) = 1$,we get:$t = \frac{2.303}{0.0693} \approx 33.23 \ min$.The nearest integer value is $33 \ min$.

If benzene diazonium chloride undergoes first order decomposition at $T \ K$ with a rate constant of $6.93 \times 10^{-2} \ min^{-1}$,the time for completion of $90 \%$ of the reaction (in $min$) is (nearest integer) $(\log 2 = 0.30, \log 3 = 0.477)$.

Similar Questions

What is the unit of the rate constant for a first-order reaction?

The first order reaction takes $20 \ min$ to complete $15\%$. How much time is required to complete $75\%$ of the reaction (in $min$)?

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

In a first order reaction,the concentration of the reactant decreases from $0.6 \ M$ to $0.3 \ M$ in $15 \ min$. The time taken for the concentration to change from $0.1 \ M$ to $0.025 \ M$ in minutes is

In a first order reaction,the concentration of the reactant decreases from $20 \ mmol \ dm^{-3}$ to $8 \ mmol \ dm^{-3}$ in $40 \ minute$. Find the rate constant of the reaction.

Vedclass Test Series

Exam Paper Generator

Online Exam Module