Rate of reaction becomes 2.5 times when temperatureis increased from 300 K to 310 K. If rate constant at300 K is K then rate constant at 310 K is

Let's solve the problem step by step:

1. We are given that the rate of reaction becomes 2.5 times when the temperature is increased from 300 K to 310 K.

2. The rate of a reaction is directly proportional to the rate constant. So, we can write the equation as:

   Rate1/Rate2 = k1/k2

   Where Rate1 and Rate2 are the rates of reaction at temperatures 300 K and 310 K respectively, and k1 and k2 are the rate constants at those temperatures.

3. We are given that Rate1/Rate2 = 2.5. Substituting this value into the equation, we get:

   2.5 = k1/k2

4. We are also given that the rate constant at 300 K is K. So, we can substitute k1 = K into the equation:

   2.5 = K/k2

5. Now, we need to solve for k2. We can rearrange the equation to isolate k2:

   k2 = K/2.5

6. Simplifying the expression, we get:

   k2 = 0.4K

7. Therefore, the rate constant at 310 K is 0.4 times the rate constant at 300 K.