Given:
- The rate of reaction A doubles on increasing the temperature from 300 K to 310 K.
- The activation energy of reaction B is twice that of reaction A.

To Find:
- By how much should the temperature of reaction B be increased from 300 K so that the rate doubles.

Explanation:
Step 1: Understanding the Effect of Temperature on Reaction Rate
- The rate of a reaction generally increases with an increase in temperature.
- This is because an increase in temperature provides more energy to the reactant molecules, which leads to more frequent and energetic collisions, and hence a higher rate of reaction.

Step 2: Effect of Temperature on Rate Constant
- The rate of a reaction can be expressed using the Arrhenius equation:
k = Ae^(-Ea/RT)
where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
- As temperature increases, the value of RT in the denominator decreases, leading to an increase in the rate constant and hence the rate of reaction.

Step 3: Relationship between Rate and Temperature
- The rate of a reaction is directly proportional to the rate constant:
rate = k[A]^m[B]^n
- If the rate constant doubles, the rate of reaction also doubles.

Step 4: Solution
- In this question, the rate of reaction A doubles on increasing the temperature from 300 K to 310 K.
- Let's assume that the rate constant of reaction A at 300 K is k1 and the rate constant at 310 K is k2.
- Since the rate doubles, we can write:
k2 = 2k1
- Now, let's consider reaction B. The activation energy of reaction B is twice that of reaction A.
- This means that the value of Ea for reaction B is 2 times the value of Ea for reaction A.
- We can use the Arrhenius equation to write the relationship between the rate constants of reaction A and reaction B:
k2 = Ae^(-Ea2/RT)
k1 = Ae^(-Ea1/RT)
- Dividing these two equations, we get:
k2/k1 = e^((Ea1-Ea2)/RT)
- Since k2 = 2k1, we can substitute this value:
2 = e^((Ea1-Ea2)/RT)
- Taking natural logarithm on both sides:
ln(2) = (Ea1-Ea2)/RT
- Rearranging the equation and substituting the given values:
Ea1 - Ea2 = ln(2) * RT
Ea1 - 2Ea1 = ln(2) * RT
Ea1 = -ln(2) * RT
- The temperature of reaction B should be increased by the same amount as the given increase in temperature for reaction A.
- Therefore, the temperature of reaction B should be increased by 10 K.
- Hence, the correct answer is option B) 4.92 K.