The activation energy for the hydrolysis of #CH_3Cl# is #"+115.9 kJ/mol"#.
Like the other answer suggested, you need to use the Arrhenius equation to determine the activation energy for this reaction.
#k =A * e^((-E_a/(RT))#, where
#k# - the rate constant of the reaction;
#A# - the pre-exponential factor;
#E_a# - the activation energy;
#R# - the universal gas constant - you'll need to use the one given in #"J/mol K"# #-># #"8.314 J/mol K"# to be exact;
#T# - the temperature in Kelvin.
This equation allows you easily calculate the activation energy when you have two values for the rate constant measured at two different temperatures. Let's take #k_1# measured at #T_1#
#k_1 = A * e^((-E_a)/(RT_1))#
and #k_2# measured at #T_2#
#k_2 = A * e^((-E_a)/(RT_2))#
Take the natural log on both sides for both of these equations, you'll get
#ln(k_1) = ln(A) + (-E_a)/(RT_1)# (1) and #ln(k_2) = ln(A) + (-E_a)/(RT_2)# (2)
We should find a way to remove #ln(A)# from the equations since no value for #A# is given to you; the easiest way to do so is to substrat equation (2) from equation (1)
#ln(k_1) - ln(k_2) = ln(A) - E_a/(RT_1) - ln(A) + E_a/(RT_2)#
You'll end up with
#ln(k_1) - ln(k_2) = -E_a/(RT_1) + E_a/(RT_2)#, which equals
#ln(k_1/k_2) = E_a/R * (1/T_2 - 1/T_1)#
Now just plug in your values and solve for #E_a# - do not forget to use temperature in Kelvin
#ln((3.32 * 10^(-10))/(3.13 * 10^(-9))) = E_a/R * (1/313.15 - 1/298.15)#
#E_a = (2.244 * "8.314 J/mol")/(0.000161) = "115.9 kJ/mol"#
