The decomposition of A into product has value of k as 4.5 x 103 s-1 at 10°C and energy of activation 60 kJ mol-1. At what temperature would k be 1.5 x 104 s-1?
Arrhenius equation, we know that,
log \(\frac{k_2}{k_1}\) = \(\frac{E_a}{2.303R}\) \(\Big(\frac{T_2 - T_1}{T_1T_2}\Big)\)
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 \(\frac{1.5 \times 10^4}{4.5 \times 10^3}\)
= \(\frac{6.0 \times 10^4 \ J mol^{-1}}{2.303 \times 8.314 \ J \ K^{-1} mol^{-1}}\)\(\Big(\frac{T_2 - 283}{283T_2}\Big)\)
⇒ 0.5229 = 3133.627\(\Big(\frac{T_2 - 283}{283T_2}\Big)\)
⇒ \(\frac{0.5229 \times 283 T_2}{3133.627}\) = T2 - 283
⇒ 0.0472 T2 = T2 - 283
⇒ 0.0472 T2 = 283
⇒ T2 = 297.019 K(approximately)
= 297 K
= 24°C
Hence, k would be 1.5 × 104 s-1 at 24°C.