The decomposition of A into product has value of

Solution:

From Arrhenius equation, we obtain

$\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$

Also, k1 = 4.5 × 103 s−1

T1 = 273 + 10 = 283 K

k2 = 1.5 × 104 s−1

Ea = 60 kJ mol−1 = 6.0 × 104 J mol−1

Then,

$\log \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}}=\frac{6.0 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}\left(\frac{T_{2}-283}{283 T_{2}}\right)$

$\Rightarrow 0.5229=3133.627\left(\frac{T_{2}-283}{283 T_{2}}\right)$

$\Rightarrow \frac{0.5229 \times 283 T_{2}}{3133.627}=T_{2}-283$

$\Rightarrow 0.0472 T_{2}=T_{2}-283$

$\Rightarrow 0.9528 T_{2}=283$

$\Rightarrow T_{2}=297.019 \mathrm{~K}$ (approximately)

= 297 K

= 24°C

Hence, k would be 1.5 × 104 s−1 at 24°C.

Note: There is a slight variation in this answer and the one given in the NCERT textbook.