For the reaction 2A + B rightarrow text{Produ | vedclass

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

Question

(A) $1$. The half-life of a reaction is independent of the initial concentration for a first-order reaction. Since the half-life remains unchanged whe

Vedclass · Accepted Answer

$L \, mol^{-1} \, s^{-1}$

Vedclass · Answer

(A) $1$. The half-life of a reaction is independent of the initial concentration for a first-order reaction. Since the half-life remains unchanged when the concentration of $A$ is doubled,the order of reaction with respect to $A$ is $1$.$2$. When the concentration of $B$ is doubled,the rate increases by a factor of $2$. This indicates that the order of reaction with respect to $B$ is $1$.$3$. The overall order of the reaction is $n = 1 + 1 = 2$.$4$. The unit of the rate constant for a reaction of order $n$ is $(mol \, L^{-1})^{1-n} \, s^{-1}$.$5$. For $n = 2$,the unit is $(mol \, L^{-1})^{1-2} \, s^{-1} = (mol \, L^{-1})^{-1} \, s^{-1} = L \, mol^{-1} \, s^{-1}$.

Rate $(mol \ L^{-1} \ s^{-1})$	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$
$0.10$	$20$	$0.5$
$0.40$	$x$	$0.5$
$0.80$	$40$	$y$

For the reaction $2A + B \rightarrow \text{Product}$,the half-life remains unchanged when the concentration of $A$ is doubled. When the concentration of $B$ is doubled,the rate of reaction increases by a factor of $2$. What is the unit of the rate constant for this reaction?

Similar Questions

For the reaction $A + B \longrightarrow \text{product}$,the rate law equation is $\text{rate} = k[A]^2[B]$. If the rate of reaction is $0.22 \ mol \ L^{-1} \ s^{-1}$,calculate the rate constant $k$. Given: $[A] = 1 \ mol \ L^{-1}, [B] = 0.25 \ mol \ L^{-1}$.

$A$ reaction involving different reactants can never be a/an......?

For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.
Rate $(mol \ L^{-1} \ s^{-1})$ $[A]$ $(mol \ L^{-1})$ $[B]$ $(mol \ L^{-1})$
$0.10$ $20$ $0.5$
$0.40$ $x$ $0.5$
$0.80$ $40$ $y$

What is the value of $x$ and $y$?

For a reaction between substances $A$ and $B$,the rate law is given by: $\text{Rate} = K[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what will be the ratio of the new rate to the original rate?

The order of a reaction which has the rate expression $\frac{dc}{dt} = K[E]^{3/2}[D]^{3/2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module