Identify the correct statement regarding the | vedclass

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

Question

(C) Step $1$: For a zero order reaction, the rate is constant and independent of the initial concentration, so $(A)$ is incorrect.Step $2$: The decomp

Vedclass · Accepted Answer

Half-life of a first order reaction is independent of the initial concentration of the reactant.

Vedclass · Answer

(C) Step $1$: For a zero order reaction, the rate is constant and independent of the initial concentration, so $(A)$ is incorrect.Step $2$: The decomposition of acetaldehyde $(CH_3CHO \rightarrow CH_4 + CO)$ is a second order reaction, so $(B)$ is incorrect.Step $3$: For a first order reaction, the half-life is given by $t_{1/2} = \frac{0.693}{k}$, which is independent of the initial concentration $[R]_0$. Thus, $(C)$ is correct.Step $4$: For a zero order reaction, the half-life is $t_{1/2} = \frac{[R]_0}{2k}$, which depends on the initial concentration. Thus, $(D)$ is incorrect.

Expt. no.	$[A]$	$[B]$	Rate $(\text{mol} \ \text{dm}^{-3} \ \text{s}^{-1})$
$1$	$0.2$	$0.2$	$2$
$2$	$0.2$	$0.4$	$4$
$3$	$0.6$	$0.4$	$36$

Identify the correct statement regarding the order of reaction from the following.

Similar Questions

Write the general reaction,its differential rate equation,and the rate law.

In the presence of $HCl$,sucrose gets hydrolysed into glucose and fructose. The concentration of sucrose was found to reduce from $0.4 \ M$ to $0.2 \ M$ in $1 \ hour$ and to $0.1 \ M$ in $2 \ hours$. The order of the reaction is:

For a reaction,$A + B \rightarrow$ products,the rate of the reaction at various concentrations is given below. The rate law for the above reaction is:
Expt. no. $[A]$ $[B]$ Rate $(\text{mol} \ \text{dm}^{-3} \ \text{s}^{-1})$
$1$ $0.2$ $0.2$ $2$
$2$ $0.2$ $0.4$ $4$
$3$ $0.6$ $0.4$ $36$

For the reaction $AB_5 \to AB + 4B$,if $- \frac{d[AB_5]}{dt} = K_1[AB_5]$ and $\frac{d[B]}{dt} = K_2[AB_5]$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module