Question

The temperature coefficient for a certain reaction is $ 2 $ . If the temperature of the reaction is raised by $ 40 $ $ {\rm K} $ , what is the approximate factor to which the rate is multiplied?

Answer 1

Hint: Arrhenius describes the mathematical relationship between the rate constant of the chemical reaction with its temperature. The Arrhenius equation is expressed as: $ k = A{e^{\dfrac{{ - Ea}}{{RT}}}} $ . Collision theory is also used to better describe the theory of Arrhenius.

Complete answer:
It is observed that temperature has a great impact over the rate of chemical reaction. Generally, it is observed that with increase in the temperature of reaction, rate of chemical reaction also increases and on decreasing the temperature the rate of reaction also decreases.
After observing such an effect, a general approximate rule was described to relate the temperature with the rate of chemical reaction. According to this rule, the rate of reaction becomes double of its initial value for a rise of $ {10^ \circ }C $ in temperature.
Temperature coefficient of the reaction is defined as the ratio of rate constant differs by $ {10^ \circ }C $ in temperature.
Temperature coefficient $ Q $ $ = $ $ \dfrac{{{k_2}{{\left( {T + 10} \right)}^ \circ }C}}{{{k_1}\left( {{T^ \circ }C} \right)}} $
Temperature coefficient is also expressed in terms of $ \left( {\log } \right) $ as:
 $ \log Q $ $ = $ $ \log $ $ \dfrac{{{k_2}{{\left( {T + 10} \right)}^ \circ }C}}{{{k_1}\left( {{T^ \circ }C} \right)}} $
Temperature coefficient of the reaction given is $ 2 $
Let the initial temperature of the reaction be $ {0^ \circ }C $ , then the value of $ T $ in the numerator becomes zero.
Temperature raised for the reaction is $ 40 $ $ {\rm K} $
 $ \log 2 $ $ = $ $ \log $ $ \dfrac{{{k_2}{{10}^ \circ }C}}{{{k_1}\left( {{{40}^ \circ }C} \right)}} $
 $ \log 2 $ $ = $ $ \dfrac{1}{4}\log \dfrac{{{k_2}}}{{{k_1}}} $
On solving the above equation on base $ 10 $ , we get
 $ \dfrac{{{k_2}}}{{{k_1}}} = {10^{4\log 2}} $
On solving the above equation we get
 $ \dfrac{{{k_2}}}{{{k_1}}} = 16 $
On rearranging the above equation, we get
 $ {k_2} = 16 \times {k_1} $
Hence, the rate of reaction will increase by $ 16 $ times the initial rate of increasing the temperature of reaction by $ 40 $ $ {\rm K} $.

Note:
The rate of reaction increases with rise in temperature because of increase in kinetic energy of molecules due to which they tend to collide with each other quickly and convert into products. Example: the process of rancidification in butter increases as the temperature rises in summer.