CHEMICAL KINETICS-ACTIVATION ENERGY DATA from the rate law experiment did you se
ID: 570374 • Letter: C
Question
CHEMICAL KINETICS-ACTIVATION ENERGY DATA from the rate law experiment did you select? Which recipe Temp.(C) Reaction time (in seconds) Temp. (K)1/Temp. (K1) In(Reaction time) iet atlust Value of activation energy from slope of graph (including units) Show calculation (d uut Reaction time with copper() ions present caleculate the activation enerey for your reaction in the presence of the copper(l)ion. Assume that the temperature did not change between the reactions with and without copper ion Cuitly antelustExplanation / Answer
since higher the temperature, lower the reaction time and higher the rate constant. the equation is
K= Ko*e(-Ea/RT)
lnK= lnKo-Ea/RT
so a plot of lnK vs 1/T( T in K) gives straight line whose slope is -Ea/R. so plot of ln(1//reaction time) ( higher the reaction time lower the rate constant and lower the reaction time, higher the reaction rate constant).So a plot of ln(1/reaction time) vs 1/T (T in K) gives straight line whose slope is -EA/R. the plot is shown below.
from the slope -Ea/R = - 6584
Ea= activation energy
R= gas constant (8.314 joules/molkelvin)
therefore Ea = activation energy in the absence of catalyst= 6584 X 8.314 joules/ mole
= 54739 joules/mole
54.739 kjoules /mole
in the presence of catalyst
let the activation energy be Ea1
arheneus equation for this case can be written as ln1/reaction time = lnKo- Ea1/RT
hence reaction time has to be compared with that of no catalyst at the same temperature for calculating activation energy in the presence of catalyst.
we assume the frequency factor Ko to be same for both the cases.Let K1 = 1/reaction time at 23 degerees centigrade( in the absence of catalyst) = 1/100.95
= 0.0099
in the presence of catalyst K2= 1/ reaction time = 1/ 29.11
= 0.034
arhenius euation in the absence and presence of catalyst can be written as lnK1= lnKo-Ea/RT ......(1)
lnK2= lnKo-Ea1/RT ......(2)
euation (2) -(1)
lnK2/lnK1= (Ea-Ea1)/RT
Ea1= Ea- RT lnK2/K1
54739- 8.314( 23+273.15)ln(0.034/0.0099)
=51701 joules/mole or 51.701 Kjoules/mole