ChemistryCBSE Class 12 · Board exam 2025, set 56/7/1 · 1 mark · MCQ

CBSE Class 12 Chemistry 2025: question 4

Set in the CBSE Class 12 board exam of 2025, paper 56/7/1, section A, question 4, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q4 · 1 mark · MCQChemical Kinetics

The activation energy ( $E_{a}$ ) of a reaction can be determined from the slope of which of the following plots ?

(A) $l n k$ vs. T
(B) $\frac{l n k}{T}$ vs. T
(C) $l n k$ vs. $\frac{1}{T}$
(D) $\frac{T}{l n k}$ vs. $\frac{1}{T}$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 56/7/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Take the logarithm of the Arrhenius equation.

Hint 2

Compare the result with the straight-line form $y = c + m x$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

The Arrhenius equation gives $ln k = ln A - \frac{E _{a}}{R} \cdot \frac{1}{T}$ .
A plot of $ln k$ vs $1/ T$ is a straight line with slope $- E_{a} / R$ , from which $E_{a}$ is found.
1 mark

Answer: (C) $ln k$ vs $1/ T$

Where the marks are

Slope $= - E_{a} / R$ from the $ln k$ vs $1/ T$ plot (1 mark)

Revise the chapter

Chapter: Chemical Kinetics (notes, formulas, where marks are lost and every past question)
Topic: Arrhenius Equation and Activation Energy

Practise this chapterBoard-style questions from Chemical Kinetics, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 56/7/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Chemistry 2025: question 4

Hints

Step-by-step solution

Revise the chapter

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