the rate of a reaction increases by 4 times when the temperature of the reaction is raised from 340 Kelvin to 360 Kelvin calculate the energy of activation of the reaction given R = 8.314J/K/mol
Answers
Answered by
33
log (k2/k1)=Ea/2.303R (T2-T1/T1.T2)
here k2/k1=4,R=8.314,T1=340,T2=360
putting these values
log4=Ea/2.303×8.314 (360-340/360×340)
2log2=Ea×20/2.303×8.314×340×360
2×0.3010=Ea×20/2343610.18
0.602=Ea/0.00000853
Ea=0.00000514
here k2/k1=4,R=8.314,T1=340,T2=360
putting these values
log4=Ea/2.303×8.314 (360-340/360×340)
2log2=Ea×20/2.303×8.314×340×360
2×0.3010=Ea×20/2343610.18
0.602=Ea/0.00000853
Ea=0.00000514
poojavijay:
sorry but the answer given is 70554J
Answered by
28
Answer:
Explanation:
Where, = rate constant at temperature
= rate constant at temperature
= activation energy
R= gas constant= 8.314J/Kmol
= 340K
= 360K
if =k, thus =4k
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