Chemistry, asked by worldfamous273, 1 day ago

6. The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature. [R = 8.314 JK-1 mol-1, log 4 = 0.6021] 52800 J mol-1 84600 J mol-1 O 52.8 K Cal mol-1 O 152.8 KJ mol-1​

Answers

Answered by sanjoosingh1182
0

Answer:

The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E

2

) of the reaction assuming that it does not change with temperature.

[R=8.314 J/K mol

−1

,log4=0.6021]log

log K

log K 1

log K 1

log K 1

log K 1 K

log K 1 K 2

log K 1 K 2

log K 1 K 2

log K 1 K 2

log K 1 K 2 =

log K 1 K 2 = 2.303R

log K 1 K 2 = 2.303RE

log K 1 K 2 = 2.303RE a

log K 1 K 2 = 2.303RE a

log K 1 K 2 = 2.303RE a

log K 1 K 2 = 2.303RE a

log K 1 K 2 = 2.303RE a (

log K 1 K 2 = 2.303RE a ( T

log K 1 K 2 = 2.303RE a ( T 1

log K 1 K 2 = 2.303RE a ( T 1

log K 1 K 2 = 2.303RE a ( T 1

log K 1 K 2 = 2.303RE a ( T 1 1

log K 1 K 2 = 2.303RE a ( T 1 1

log K 1 K 2 = 2.303RE a ( T 1 1 −

log K 1 K 2 = 2.303RE a ( T 1 1 − T

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4=

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a (

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 −

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a =0.6021×2.303×8.314×293×

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a =0.6021×2.303×8.314×293× 20

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a =0.6021×2.303×8.314×293× 20313

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a =0.6021×2.303×8.314×293× 20313

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a =0.6021×2.303×8.314×293× 20313

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a =0.6021×2.303×8.314×293× 20313 E

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a =0.6021×2.303×8.314×293× 20313 E a

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a =0.6021×2.303×8.314×293× 20313 E a

log K 1 K 2 = 2.303RE a ( T 1 1 − T 2 1 )log4= 2.303×8.314E a ( T 1 1 − T 2 1 )E a =0.6021×2.303×8.314×293× 20313 E a =52.863kJ

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