
The rate of a reaction is doubled for every $ {10^o}C $ rise in temperature. The increase in rate as result of increase in temperature from $ {10^o}C $ to $ {100^o}C $ is:
$ (A)112 $
$ (B)512 $
$ (C)400 $
$ (D)256 $
Answer
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Hint: In chemical kinetics, according to the Arrhenius equation, we can understand that the rate of a reaction is directly proportional to the temperature conditions. We are also given that for every $ {10^o}C $ rise in temperature, the rate of a reaction gets doubled. So, based on this information we will find the increase in rate due to the increase in temperature from $ {10^o}C $ to $ {100^o}C $ .
Complete Step By Step Answer:
We will first see the Arrhenius equation which shows that the rate of a reaction is directly proportional to the temperature.
$ \ln k = - \dfrac{{{E_a}}}{{RT}} + \ln A $
Where k is the rate constant for the reaction, $ {E_a} $ is energy of activation, R is known as the universal gas constant, T is the temperature represented in Kelvin and A is known as the pre-exponential factor.
So, for every $ {10^o}C $ rise in temperature, the rate of a reaction gets doubled. Therefore, the rate of reaction changes by $ {2^n} $ times.
Where, n is the number of times the temperature is increased by $ {10^o}C $ .
Therefore, the value of n when temperature increases from $ {10^o}C $ to $ {100^o}C $ is $ 9. $
Hence, the new rate of reaction is $ {2^9}. $
The new rate of reaction is $ 512. $
Therefore, the correct option is $ (B)512. $ .
Note:
The rate of a reaction increases with the increase in temperature because collision between the molecules of the reactants increases. This happens because on increasing temperature, the kinetic energy of the molecules also increases. The ratio of the rate constant at two temperatures with a difference of ten degrees Celsius is known as temperature coefficient.
Complete Step By Step Answer:
We will first see the Arrhenius equation which shows that the rate of a reaction is directly proportional to the temperature.
$ \ln k = - \dfrac{{{E_a}}}{{RT}} + \ln A $
Where k is the rate constant for the reaction, $ {E_a} $ is energy of activation, R is known as the universal gas constant, T is the temperature represented in Kelvin and A is known as the pre-exponential factor.
So, for every $ {10^o}C $ rise in temperature, the rate of a reaction gets doubled. Therefore, the rate of reaction changes by $ {2^n} $ times.
Where, n is the number of times the temperature is increased by $ {10^o}C $ .
Therefore, the value of n when temperature increases from $ {10^o}C $ to $ {100^o}C $ is $ 9. $
Hence, the new rate of reaction is $ {2^9}. $
The new rate of reaction is $ 512. $
Therefore, the correct option is $ (B)512. $ .
Note:
The rate of a reaction increases with the increase in temperature because collision between the molecules of the reactants increases. This happens because on increasing temperature, the kinetic energy of the molecules also increases. The ratio of the rate constant at two temperatures with a difference of ten degrees Celsius is known as temperature coefficient.
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