The rate equation is given as:
k = A e^(- Ea / RT)
Dividing state 1 and state 2:
k1/k2 = e^(- Ea / RT1) / e^(- Ea / RT2)
k1/k2 = e^[- Ea / RT1 - (- Ea / RT2)]
k1/k2 = e^[- Ea / RT1 + Ea / RT2)]
Taking the ln of both sides:
ln (k1/k2) = - Ea / RT1 + Ea / RT2
ln (k1/k2) = - Ea / R (1/T1 - 1/T2)
Since k2 = 4k1, therefore k1/k2 = ¼
ln (1/4) = [- (56,000 J/mol) / (8.314 J / mol K)] (1/273
K – 1/ T2)
2.058 x 10^-4 = 1/273 – 1/T2
T2 = 289.25 K
A boat being accelerated by the engine
Bringing water to the boil in an electric kettle
Answer:
Explanation:
Pair 2.50g of O₂ and 2.50g of N₂
The atoms sample with the largest number of moles since the masses are the same would be the one with lowest molar mass according the the equation below:
Number of moles = 
Atomic mass of O = 16g and N = 14g
Molar mass of O₂ = 16 x 2 = 32gmol⁻¹
Molar mass of N₂ = 14 x 2 = 28gmol⁻¹
Number of moles of O₂ =
= 0.078mole
Number of moles of N₂ =
= 0.089mole
We see that N₂ has the largest number of moles