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Chemistry · 2026 · 3 marks
CBSE 2026 · Region 5 · Set 1 · Q24
The rate of the chemical reaction doubles when the temperature is raised from 298 K to 308 K. Calculate activation energy ( E_a ) for this reaction…
The rate of the chemical reaction doubles when the temperature is raised from $\displaystyle 298$ K to $\displaystyle 308$ K. Calculate activation energy $\displaystyle \left(\mathrm{E}_{\mathrm{a}}\right)$ for this reaction assuming that it does not change with temperature. (Given : $\displaystyle \mathrm{R}=8 \cdot 314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 2=0 \cdot 30$ )
Marking-scheme solution
K
a
log K
2.303R
T
T
(
)
=
−
(
)
(
)
E
K
K
=
∴
Ea
log2
(
)
=
−
(
)
×
(
)
$\displaystyle 0.30$ =
Ea
𝐸𝑎= $\displaystyle 0.30$ × $\displaystyle 19.147$ × $\displaystyle 298$ × $\displaystyle 308$
= $\displaystyle 52722$ J mol−$\displaystyle 1$ or = $\displaystyle 52.722$ kJ mol−$\displaystyle 1$ .
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