# The rate constant for a reaction at 25.0 degrees C is 0.010 #s^-1# and its activation energy is 35.8 KJ. How do you find the rate constant at 50.0 degrees C?

##### 1 Answer

You compare them at two temperatures and determine the Arrhenius plot.

#color(white)(=>)k_2 = k_1e^((E_a)/R[1/(T_1) - 1/(T_2)])#

Be sure to use

You should get

The Arrhenius equations are:

#k_1 = Ae^(-E_a"/"RT_1)#

#k_2 = Ae^(-E_a"/"RT_2)#

The **same** reaction has the **same** activation energy **same** universal gas constant

Now, divide these equations.

#k_1/k_2 = e^(-E_a"/"RT_1)/e^(-E_a"/"RT_2)#

taking the natural logarithm allows you to separate the right side.

#ln(k_1/k_2) = ln(e^(-E_a"/"RT_1)/e^(-E_a"/"RT_2))#

#= ln(e^(-E_a"/"RT_1)) - ln(e^(-E_a"/"RT_2))#

#= -E_a/(RT_1) - (-E_a)/(RT_2)#

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

So your final equation is:

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

To find the rate constant at a new temperature, try switching the sign on the left and right:

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

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

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

There, now it's easier to solve for

#k_2/k_1 = e^((E_a)/R[1/(T_1) - 1/(T_2)])#