The reaction L rightarrow M starts with 10 , | 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 constant $k$ is given by $k = \frac{2.303}{t} \log(\frac{[A]_0}{[A]_t})$.At $t_1 = 30 \, min$,$[A]_0 = 10 \, g

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log(\frac{[A]_0}{[A]_t})$.At $t_1 = 30 \, min$,$[A]_0 = 10 \, gL^{-1}$ and $[A]_t = 5 \, gL^{-1}$.$k = \frac{2.303}{30} \log(\frac{10}{5}) = \frac{2.303}{30} \log(2)$.At $t_2 = 90 \, min$,$[A]_0 = 10 \, gL^{-1}$ and $[A]_t = 1.25 \, gL^{-1}$.$k = \frac{2.303}{90} \log(\frac{10}{1.25}) = \frac{2.303}{90} \log(8) = \frac{2.303}{90} \log(2^3) = \frac{2.303 	imes 3}{90} \log(2) = \frac{2.303}{30} \log(2)$.Since the rate constant $k$ is the same in both cases,the reaction is of the first order.

The reaction $L \rightarrow M$ starts with $10 \, gL^{-1}$. After $30$ and $90$ minutes,$5 \, gL^{-1}$ and $1.25 \, gL^{-1}$ remain,respectively. The order of the reaction is:

Similar Questions

Thermal decomposition of $HCOOH$ is a first order reaction and the rate constant at $T(K)$ is $4.606 \times 10^{-3} \ s^{-1}$. The time required to decompose $90 \%$ of initial quantity of $HCOOH$ at $T(K)$ in seconds is

If the decomposition reaction $A_{(g)} \to B_{(g)}$ follows first-order kinetics,then the graph of the rate of formation $(R)$ of $B$ against time $t$ will be:

If a $75\%$ first-order reaction is completed in $32 \text{ minutes}$,how many minutes will it take for the same reaction to be $50\%$ completed?

Consider the first order reaction $R \to P$. The fraction of molecules decomposed in the given first order reaction can be expressed as:

For a first-order reaction,the concentration decreases from $800 \ mol/dm^3$ to $50 \ mol/dm^3$ in $2 \times 10^4 \ s$. What is the rate constant of the reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module