The activation energy of a certain reaction is 46.6 kj/mol. At 20 degrees Celsius the rate constant is #0.0130s^-1#. At what temperature would this reaction go twice as fast (T2)?

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Michael Share
Mar 20, 2016

Answer:

At #31^@"C"#

Explanation:

We can use The Arrhenius Equation:

#k=Ae^(-E_a/(RT))#

#k# is the rate constant

#A# is the frequency factor and is constant

#R# is the gas constant and #= 8.31"J""/""K""/""mol"#

#E_a# is the activation energy

#T# is the absolute temperature.

A more usable form of the equation is obtained by taking natural logs of both sides #rArr#

#lnk=lnA-E_a/(RT)#

If the reaction occurs at two temperatures #T_1# and #T_2# then we can write:

#lnk_1=lnA-E_a/(RT_1)" "color(red)((1))#

#lnk_2=lnA-E_a/(RT_2)" "color(red)((2))#

Now we can subtract both sides of #color(red)((1))# and #color(red)((2))rArr#

#ln(k_1/k_2)=-E_a/(RT_1)+E_a/(RT_2)#

Since we know that #k_2=2k_1# this becomes:

#ln(1/2)=E_a/R[1/T_2-1/T_1]#

#:.-0.693=(46.6xx10^3)/ 8.31[1/T_2-1/293]#

#:.1/T_2-1/293=-1.235xx10^-4#

#:.1/T_2=-1.245xx10^-4+34.13xx10^-4=34.13xx10^-4#

#:.T_2=304color(white)x"K"#

#:.T_2=304-273=34^@"C"#

This shows how a rise of just 14 degrees can double the rate of reaction.

The graphic below shows how a small increase in temperature results in a large increase in the number of particles which have the minimum energy for a reaction to happen:

chemguide.co.uk

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