Thermal decomposition of a compound is of fir | vedclass

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

Question

(C) For a first-order reaction,the half-life is given by $t_{1/2} = 120 \ min$.The rate constant $K$ is calculated as: $K = \frac{0.693}{t_{1/2}} = \f

Vedclass · Accepted Answer

$398.8$

Vedclass · Answer

(C) For a first-order reaction,the half-life is given by $t_{1/2} = 120 \ min$.The rate constant $K$ is calculated as: $K = \frac{0.693}{t_{1/2}} = \frac{0.693}{120} = 5.775 	imes 10^{-3} \ min^{-1}$.For $90\%$ decomposition,the amount remaining is $10\%$ of the initial concentration $([A]_0 = 100, [A]_t = 10)$.Using the first-order rate equation: $t = \frac{2.303}{K} \log \frac{[A]_0}{[A]_t}$.$t_{90\%} = \frac{2.303}{5.775 	imes 10^{-3}} \log \frac{100}{10} = \frac{2.303}{5.775 	imes 10^{-3}} 	imes 1 = 398.8 \ min$.

Thermal decomposition of a compound is of first order. If $50\%$ of a sample of this compound is decomposed in $120 \ min$,then how long will it take $90\%$ of the compound to decompose? ........ $min.$

Similar Questions

$A$ 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?

For a first-order reaction involving gaseous reactants and gaseous products,what will be the unit of its rate constant?

The half-life of a first-order reaction is $1 \text{ hour}$. What fraction of the reactant will remain after $3 \text{ hours}$?

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

In a first-order reaction,the concentration of the reactant decreases from $1.0 \, M$ to $0.25 \, M$ in $20 \, \text{minutes}$. What will be the rate constant of the reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module