The decomposition of A into product has value of k as 4.5 x 10^3 s^1 at 10°C and energy of activation 60 kJ mol^1. At what temperature would k be 1.5 x 10^4 s^1?
The decomposition of A into product has value of k as 4.5 x 10^{3} s^{1} at 10°C and energy of activation 60 kJ mol^{1}. At what temperature would k be 1.5 x 10^{4} 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, k_{1} = 4.5 × 10^{3} s^{1}
T_{1} = 273 + 10 = 283 K
k_{2} = 1.5 × 10^{4} s^{1}
E_{a} = 60 kJ mol^{1} = 6.0 × 10^{4} 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\(\Big(\frac{T_2  283}{283T_2}\Big)\)
⇒ \(\frac{0.5229 \times 283 T_2}{3133.627}\) = T_{2}  283
⇒ 0.0472 T_{2} = T_{2}  283
⇒ 0.0472 T_{2} = 283
⇒ T_{2} = 297.019 K(approximately)
= 297 K
= 24°C
Hence, k would be 1.5 × 10^{4} s^{1} at 24°C.

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 x 10^10 s^1. Calculate k at 318 K and Ea.
2 years ago

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
2 years ago

The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34  1.25 x 10^4 K/T.
2 years ago

The decomposition of hydrocarbon follows the equation k = (4.5 x 10^11 s^1) e^28000 K/T Calculate Ea.
2 years ago

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
2 years ago
 321 Forums
 27.3 K Topics
 53.8 K Posts
 0 Online
 12.4 K Members