Afvak A.
asked • 02/09/22How to find activation energy (kJ/mol) given the following?
The equation of a trendline of 1/Temp (K^-1) vs. ln(1/time) --> y = -4548.6x + 10.905
R = 8.314 J / k mol
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1 Expert Answer
Corban E. answered • 02/09/22
Tutor
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(401)
AP Chemistry Tutor and Former Teacher (Gen Chem, IB, O-Level, A-Level)
y =(-4548.6x)+10.905
When you get these, it's
y axis=ln k .... k is the rate constant
x axis = 1/T ... T is in Kelvin
And the slope is equal to:
slope= -Ea/R
where Ea will be in J/mol because R is 8.314 J/mol K.
slope= -4548.6
R=8.314 J/mol K.
slope= -Ea/R
-Ea= slope (R)
-Ea=(-4548.6)(8.314)
-Ea=-37817.0604 J/mol
Ea=37817.0604 J/mol
103J=1 kJ, so
Ea=37.8170604 kJ/mol
Ea=37.82 kJ/mol
J.R. S.
tutor
It would have been nice if the OP had used rate constant k , instead of temperature in K Kelvin and t for time.
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02/09/22
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J.R. S.
02/09/22