The activation energy for this reaction is equal to **+161 kJ/mol**.

You were given the *rate constant* for a reaction at two different temperatures, **298 K** and **425 K**. This means tha you can use the **Arhennius equation** to calculate the activation energy for this reaction

#k = A * e^(-E_A/(RT))#, where

#k# - the rate constant;

#A# - the *Arhennius factor*;

#E_A# - the activation energy for the reaction;

#R# - the *gas constant*, equal to #8.314 "J/mol"^(-1)"K"^(-1)#

#T# - the tmperature at which the reaction takes place - expressed in Kelvin.

For the two rate constants given, you can write

#k_1 = A * e^(-E_a/(RT_1))# and #k_2 = A * e^(-E_A/(RT_2))#

Take the natural log from both sides of the equations to get

#ln(k_1) = lnA + (-E_A/(RT_1))# #" "color(blue)((1))#

and

#ln(k_2) = lnA + (-E_A/(RT_2))# #" "color(blue)((2))#

Subtract equation #color(blue)((2))# from equation #color(blue)((1))# to get

#ln(k_1) - ln(k_2) = cancel(ln(A)) - cancel(ln(A)) - E_A/(RT_1) - (-E_A/(RT_2))#

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

Now plug your values into the equation and solve for #E_A#

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

#E_A = (ln((3.16 * 10^(-15)cancel("s"^(-1)))/(8.66 * 10^(-7)cancel("s"^(-1)))) * 8.314"J"/("mol" * cancel("K")))/((1/425 - 1/298)cancel("K")) = (-161531"J"/"mol")/((1/425 - 1/298))#

#E_A = "161086 J/mol"#

Rounded to three sig figs and expressed in kJ per mol, the answer will be

#E_A = color(green)("+161 kJ/mol")#