In a first order reaction,87.5 % of the react | vedclass

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

Question

(A) For a first order reaction,the amount of reactant remaining after $87.5 \%$ conversion is $100 \% - 87.5 \% = 12.5 \%$.Let the initial concentrati

Vedclass · Accepted Answer

$\frac{0.693}{5} \ min^{-1}$

Vedclass · Answer

(A) For a first order reaction,the amount of reactant remaining after $87.5 \%$ conversion is $100 \% - 87.5 \% = 12.5 \%$.Let the initial concentration $[A]_0 = 100$ and the concentration at time $t = 15 \ min$ be $[A]_t = 12.5$.The number of half-lives $n$ can be calculated as $12.5 = 100 	imes (1/2)^n$,which gives $(1/2)^n = 1/8$,so $n = 3$.Since $t = n 	imes t_{1/2}$,we have $15 = 3 	imes t_{1/2}$,which means $t_{1/2} = 5 \ min$.The rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5} \ min^{-1}$.

In a first order reaction,$87.5 \%$ of the reactant is converted into the product in $15 \ minutes$. The rate constant for the reaction is given by:

Similar Questions

Which is the correct relation between rate constant and half-life of a first-order reaction?

For the first-order reaction $N_2O_5 \to 2NO_2 + \frac{1}{2}O_2$,the half-life at $30 \ ^\circ C$ is $24 \ \text{hours}$. Starting with $10 \ g$ of $N_2O_5$,how many grams of $N_2O_5$ will remain after $96 \ \text{hours}$ (in $g$)?

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

When the value of rate constant $(k)$ is $2.0 \text{ min}^{-1}$,then what will be the half-life of the reaction $(t_{1/2})$ in seconds?

For a first order reaction $A \rightarrow B$,the rate constant,$k = 5.5 \times 10^{-14} \, s^{-1}$. The time required for $67 \, \%$ completion of reaction is $x \times 10^{-1}$ times the half life of reaction. The value of $x$ is $....$ (Nearest integer)

Vedclass Test Series

Exam Paper Generator

Online Exam Module