(A) For a first-order reaction,the rate law is given by: $\text{Rate} = k[\text{Reactant}]$.Comparing this with the equation of a straight line $y = m

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(A) For a first-order reaction,the rate law is given by: $\text{Rate} = k[\text{Reactant}]$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \text{Rate}$,$x = [\text{Concentration}]$,and $c = 0$,the slope $m$ is equal to the rate constant $k$.Given that the slope is $2.5 \times 10^{-3}$,therefore,the rate constant $k = 2.5 \times 10^{-3} \ s^{-1}$.

Class 12 Chemistry Chemical Kinetics First Order reaction

Medium

What is the value of the rate constant for a first-order reaction if the slope for the graph of rate versus concentration is $2.5 \times 10^{-3}$?

MHT CET 2023 All MHT CET

A
$2.5 \times 10^{-3} \ s^{-1}$
B
$5.0 \times 10^{-3} \ s^{-1}$
C
$7.5 \times 10^{-3} \ s^{-1}$
D
$1.25 \times 10^{-3} \ s^{-1}$

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What is the value of the rate constant for a first-order reaction if the slope for the graph of rate versus concentration is $2.5 \times 10^{-3}$?

Similar Questions

For the first-order reaction $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$,which of the following statements is incorrect?

For a first order reaction,$A \to P$,$t_{1/2}$ (half-life) is $10 \ days$. The time required for $\frac{1}{4}$ conversion of $A$ (in days) is: $(\ln 2 = 0.693, \ln 3 = 1.1)$.

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