1. ⇒ (MHT CET 2023 11th May Morning Shift )

Slope of the graph between $\log \frac{[A]_{0}}{[A]_{t}}$ (y axis) and time ( $x$ axis) for first order reaction is equal to:

A. $+ \frac{k}{2.303}$

B. $k$

C. $- k$

D. $- \frac{2.303}{k}$

Correct Option is (A)

The integrated rate law for the first order reaction is

$k = \frac{2.303}{t} \log_{10} \frac{[A]_{0}}{[A]_{t}}$

MHT CET 2023 11th May Morning Shift Chemistry - Chemical Kinetics Question 11 English Explanation

The graph of $\log_{10} \frac{[A]_{0}}{[A]_{t}}$ versus time $(t)$ is a straight line passing through origin with slope $(m) = + \frac{k}{2.303}$ .

2. ⇒ (MHT CET 2023 9th May Morning Shift )

Which from following is the slope of the graph of rate versus concentration of the reactant for first order reaction?

A. $- k$

B. $k$

C. $\frac{k}{2.303}$

D. $\frac{- k}{2.303}$

Correct Option is (B)

For a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant. The rate law for a first-order reaction can be written as :

$Rate = k [Reactant]$

where $k$ is the rate constant.

The graph of rate versus concentration of the reactant for a first-order reaction would be a straight line with a slope equal to the rate constant $k$ . This is because the rate is directly proportional to the concentration, making the slope of the line the rate constant itself.

Therefore, the correct answer is :

Option B : $k$

3. ⇒ (MHT CET 2021 21th September Morning Shift )

Slope of the graph between rate ( $Y$ -axis) and $[A] (X$ -axis) for the first order reaction is equal to

A. $k$

B. $\frac{2.303}{k}$

C. $\frac{k}{2.303}$

D. $- k$

Correct Option is (A)

A plot of rate versus [A] $_{t}$ is a straight line passing through origin. This is shown in Fig. The slope of straight line = k.

MHT CET 2021 21th September Morning Shift Chemistry - Chemical Kinetics Question 28 English Explanation

⇒Chemical Kinetics