The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much should the temperature of reaction B be increased from 300 K so that the rate doubles if the activation energy of reaction B is twice that of reaction A?

For reaction A, the Arrhenius equation is:

log(k'/k) = Ea / (2.303R) * [(T' - T) / (TT')]

where:
k' = rate constant at temperature T'
k = rate constant at temperature T
Ea = activation energy
R = gas constant
T' = final temperature
T = initial temperature

Given that the rate doubles when the temperature increases from 300 K to 310 K, we have k' = 2k, T = 300 K, and T' = 310 K. Substituting these values into the equation, we get:

log(2) = Ea / (2.303R) * [(310 - 300) / (300 * 310)]

log(2) = Ea / (2.303R) * [10 / 93000]

For reaction B, the activation energy is twice that of reaction A, so Ea(B) = 2Ea(A). Let's denote the temperature increase for reaction B as ΔT. Then we have:

log(2) = 2Ea(A) / (2.303R) * [ΔT / (300 * (300 + ΔT))]

Since log(2) is the same for both reactions, we can equate the two expressions:

Ea / (2.303R) * [10 / 93000] = 2Ea / (2.303R) * [ΔT / (300 * (300 + ΔT))]

Simplifying, we get:

10 / 93000 = 2ΔT / (300 * (300 + ΔT))

1 / 9300 = ΔT / (90000 + 300ΔT)

90000 + 300ΔT = 9300ΔT

90000 = 9000ΔT

ΔT = 10 K

However, this is not one of the options. Let's re-examine the derivation. The correct equation should be:

log(2) = Ea / (2.303R) * [(1/T) - (1/T')]

For reaction A:
log(2) = Ea / (2.303R) * [(1/300) - (1/310)]

For reaction B:
log(2) = 2Ea / (2.303R) * [(1/300) - (1/(300 + ΔT))]

Equating the two expressions:
(1/300) - (1/310) = 2[(1/300) - (1/(300 + ΔT))]
(1/300) - (1/310) = (2/300) - (2/(300 + ΔT))
(2/(300+ΔT)) - (1/310) = (1/300)
(2/(300 + ΔT)) = (1/300) + (1/310) = 610/93000
300 + ΔT = 2 * 93000 / 610
300 + ΔT ≈ 305
ΔT ≈ 5K

This calculation also does not align perfectly with the given options. There might be a minor error in the problem statement or the provided options.