An organic compound undergoes first-order dec | vedclass

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

Question

(B) For a first-order reaction,the rate constant $K$ is given by $K = \frac{1}{t} \ln \left( \frac{C_0}{C_t} \right)$.For $t_{1/8}$,the concentration

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) For a first-order reaction,the rate constant $K$ is given by $K = \frac{1}{t} \ln \left( \frac{C_0}{C_t} ight)$.For $t_{1/8}$,the concentration $C_t = C_0 / 8$,so $K t_{1/8} = \ln(8) = 3 \ln(2)$.For $t_{1/10}$,the concentration $C_t = C_0 / 10$,so $K t_{1/10} = \ln(10)$.Taking the ratio: $\frac{t_{1/8}}{t_{1/10}} = \frac{3 \ln(2)}{\ln(10)} = 3 \log_{10} 2$.Given $\log_{10} 2 = 0.3$,we have $\frac{t_{1/8}}{t_{1/10}} = 3 	imes 0.3 = 0.9$.Therefore,$\frac{t_{1/8}}{t_{1/10}} 	imes 10 = 0.9 	imes 10 = 9$.

An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1/8$ and $1/10$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$? (Given: $\log_{10} 2 = 0.3$)

Similar Questions

For a $1^{st}$ order reaction $R \rightarrow P$, the concentration of reactant $R$ changes from $0.1 \text{ M}$ to $0.025 \text{ M}$ in $40 \text{ minutes}$. The rate of reaction when the concentration of $R$ is $0.01 \text{ M}$ is

Which of the following represents the integrated rate law equation for the gas-phase first-order reaction,$A_{(g)} \rightarrow B_{(g)} + C_{(g)}$? If $P_{i}$ is the initial pressure of $A$ and $P$ is the total pressure of the reaction mixture at time $t$.

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

Show that in a first order reaction,the time required for completion of $99.9 \%$ is $10$ times the half-life $(t_{1/2})$ of the reaction.

For a first order reaction $A_5 \rightarrow 5 B_2$,the concentration vs time plot is as shown. The half-life of the reaction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module