The activation energy of a certain reaction is 35.3 kJ/mol. At 20 degrees C, the rate constant is #0.0130 s^-1#. At what temperature would this reaction go twice as fast? please help?

1 Answer
May 8, 2016

Answer:

#T_2=308K=35^@C#

Explanation:

To solve this question, we can modify the Arrhenius equation which provides a relationship between the equilibrium constant #k# and temperature #T#:

#lnk=-(E_a)/R(1/T)+lnA#

To find the new temperature #T_2# of the reaction in this question, we can use the following expression:

#ln((k_2)/(k_1))=(E_a)/R(1/(T_1)-1/(T_2))#

If the rate is doubled, the equilibrium constant will also double and therefore, #k_2=2xxk_1=2xx0.0130s^(-1)=0.0260s^(-1)#

Therefore, #ln((0.0260cancel(s^(-1)))/(0.0130cancel(s^(-1))))=(35300(cancel(J))/(cancel(mol)*cancel(K)))/(8.31(cancel(J))/(cancel(mol)*cancel(K)))(1/(293K)-1/(T_2))#

Solving for #T_2#: #=>T_2=308K=35^@C#

Here is a video that explains the theory behind this topic:
Chemical Kinetics | A Model for Chemical Kinetics & Catalysis.