Verified Solution • 2025–26 Session moderate

QJEEM21Ch4f19: JEE Main 2021 Chemistry - Class 12 JEEM21Ch4f19 Solution: The first order rate constant for the decomposition of CaCO 3…

The first order rate constant for the decomposition of CaCO3 at 700 K is 6.36 ×\times 10 −- 3s −- 1 and activation energy is 209 kJ mol −- 1. Its rate constant (in s −- 1) at 600 K is x ×\times 10 −- 6. The value of x is ___________. (Nearest integer)

[Given R = 8.31 J K −- mol −- 1; log 6.36 ×\times 10 −- 3 = −- 2.19, 10 −- 4.79 = 1.62 ×\times 10 −- 5]

Step-by-Step Verified Solution

K700 = 6.36 ×\times 10 −- 3s −- 1;

K600 = x ×\times 10 −- 6s −- 1

Ea = 209 kJ/mol

Applying;

log⁡(KT2KT1)=−Ea2.303R(1T2−1T1)\log \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)

log⁡(K700K600)=−Ea2.303R(1700−1600)\log \left( {{{{K_{700}}} \over {{K_{600}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {700}} - {1 \over {600}}} \right)

log⁡(6.36×10−3K600)=+209×10002.303×8.31(100700×600)\log \left( {{{6.36 \times {{10}^{ - 3}}} \over {{K_{600}}}}} \right) = {{ + 209 \times 1000} \over {2.303 \times 8.31}}\left( {{{100} \over {700 \times 600}}} \right)

log(6.36 ×\times 10 −- 3) −- logK600 = 2.6

⇒\Rightarrow logK600 = −- 2.19 −- 2.6 = −- 4.79

⇒\Rightarrow K600 = 10 −- 4.79 = 1.62 ×\times 10 −- 5

= 1.62 ×\times 10 −- 6

= x ×\times 10 −- 6

⇒\Rightarrow x = 16

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