ChemistryCBSE Class 12 · Sample paper, session 2026-27 · 3 marks · Short answer

CBSE Class 12 Chemistry sample paper 2026-27: question 23

From the official CBSE sample question paper for 2026-27, section C, question 23, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q23 · 3 marks · Short answerChemical Kinetics

The values of log k and 1/T for a reaction are given-

S.No.	Log k	1/T (/K)
1.	-4.46	0.00333
2.	-3.91	0.00323
3.	-3.37	0.00313
4.	-2.82	0.00303
5.	-2.31	0.00294

The plot of log k vs 1/T obtained for the above data is as shown in Figure 3.

Here k is the rate constant and T is the absolute temperature.

Calculate $E_{a}$ for the reaction. (Given R= 8.314 $J K^{- 1} m o l^{- 1}$ )

Figure in the paper: Figure 3: straight line graph of log k (y-axis, from -5 to 0) versus 1/T (x-axis values 0.00333 to 0.00294), rising line through points (0.00333, -4.46), (0.00323, -3.91), (0.00313, -3.37), (0.00303, -2.82), (0.00294, -2.31). See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Take log of $k = A e^{- E_{a} / R T}$ and compare it with a straight-line equation.

Hint 2

Use any two points from the table to find the slope.

Hint 3

Slope = −Ea/(2.303R). Solve for Ea and convert J to kJ.

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.

Arrhenius equation in log form: $lo g k = lo g A - \frac{E _{a}}{2.303 R T}$ .
½ mark
Comparing with $y = m x + c$ for a plot of $lo g k$ vs $1/ T$ : slope $= - \frac{E _{a}}{2.303 R}$ .
½ mark
Slope from points 1 and 2: $\frac{- 4.46 - ( - 3.91 )}{0.00333 - 0.00323} = \frac{- 0.55}{0.0001} = - 5500$ K
1 mark
$- \frac{E _{a}}{2.303 R} = - 5500$
$E_{a} = 5500 \times 2.303 \times 8.314 = 105307 J mol^{- 1} \approx 105.3 kJ mol^{- 1}$
1 mark

Answer: $E_{a} \approx 105.3$ kJ mol⁻¹ (slope = −5500 K)

Where the marks are

Arrhenius equation in log form (½ mark)
Slope = −Ea/2.303R (½ mark)
Slope ≈ −5500 K calculated from the data (1 mark)
Ea ≈ 105.3 kJ mol⁻¹ with correct unit (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 paperSample paper 2026-27: every question, by section.

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CBSE Class 12 Chemistry sample paper 2026-27: question 23

Hints

Step-by-step solution

Revise the chapter

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