Question

13. For X+Y =3. If the rate constant at 300
k is 'Q' min!. at what temperature rate
constant becomes 19Q' min '?
1) 47°C
2) 320°C
3) 280 K
4) V9x300 K​

Answer 1

Given: The correct question is: for X --> Y, $(k_{t} + 10 / k_{t} ) = 3$, rate constant at 300  k is 'Q' min^-1.

To find: At what temperature rate constant becomes '9Q' min^-1?

Solution:

• Now, we have given the rate constant as 300 k and temperature coefficient is μ = 3

• Also now:

            k(T2) / k(T1) = μ ^ (T2 - T1 / 10 )

           9Q / Q = 3 ^ (T2 - 300 / 10 )

           9 = 3 ^ (T2 - 300 / 10 )

           3 ^2 =  3 ^ (T2 - 300 / 10 )

• Now equating the powers as the base is same, we get:

           2 = T2 - 300 / 10

           20 = T2 - 300

           T2 = 320K

Answer:

             So at 320 K, rate constant becomes 9Q min^-1.