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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

Question

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

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Solution

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.

This problem has been solved

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