For a first order gas phase elementary reacti | vedclass

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

Question

(D) For a first-order reaction,the half-life is given by $t_{50\%} = \frac{\ln 2}{k}$.For $t_{87.5\%}$,the amount remaining is $100 - 87.5 = 12.5\%$,w

Vedclass · Accepted Answer

$t_{90\%} = 4 \cdot t_{50\%}$

Vedclass · Answer

(D) For a first-order reaction,the half-life is given by $t_{50\%} = \frac{\ln 2}{k}$.For $t_{87.5\%}$,the amount remaining is $100 - 87.5 = 12.5\%$,which is $(1/8)$ of the initial amount. Thus,$t_{87.5\%} = \frac{\ln 8}{k} = \frac{3 \ln 2}{k} = 3 \cdot t_{50\%}$.For $t_{90\%}$,the amount remaining is $100 - 90 = 10\%$,which is $(1/10)$ of the initial amount. Thus,$t_{90\%} = \frac{\ln 10}{k} \approx \frac{2.303}{k} \approx 3.32 \cdot t_{50\%}$.Therefore,the statement $t_{90\%} = 4 \cdot t_{50\%}$ is $INCORRECT$.

For a first order gas phase elementary reaction $A_{(g)} \xrightarrow{k} 2B_{(g)}$,which of the following is $INCORRECT$?

Similar Questions

If the concentration unit decreases by a factor of $n$,what will be the value of the rate constant for a first-order reaction?

For a first order reaction,the rate constant is $0.6932 \ hr^{-1}$,then the half-life for the reaction is ......... $hr$.

$A$ first order reaction takes $40 \ min$ for $30 \%$ decomposition. Calculate $t_{1/2}$.

For a first-order reaction,the time taken for the initial concentration to decrease to $1/4$ of its value is $20 \ min$. The time taken for the initial concentration to decrease to $1/16$ of its value will be $......... \ min$.

$A$ first order reaction is found to have a rate constant,$k = 5.5 \times 10^{-14} \ s^{-1}$. The half life of reaction is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module