Question 4.22: The rate constant for the decomposition of N2O5 at various temperatures is given below:
Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30º and 50ºC.
Answer
Take antilog both side we get
A = 2.91 x 106
At 30º C temperature in Kelvin T = 30 + 273 = 303 K
Value of 1/ T = 1/ 303 = 3.3 x 10 – 3 K-1
Plug the values in formula now we get
ln k = -2.8
make value positive to take antilog
add and subtract 3 both side we get
ln k = -2.8 + 3 -3
ln k = 0.2 - 3
Take anti ln both side we get
k =6.06 x 10-2 sec -1
Similarly calculate for 50ºC
At 50º C, K 50 + 273 = 323 K
1/T = 1/323 = 3.1 x 10 -3 K-1
Plug the values we get \
ln k = -0.5
take antiln both side we get
k = 0.607 sec -1
Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30º and 50ºC.
Answer
Take antilog both side we get
A = 2.91 x 106
At 30º C temperature in Kelvin T = 30 + 273 = 303 K
Value of 1/ T = 1/ 303 = 3.3 x 10 – 3 K-1
Plug the values in formula now we get
ln k = -2.8
make value positive to take antilog
add and subtract 3 both side we get
ln k = -2.8 + 3 -3
ln k = 0.2 - 3
Take anti ln both side we get
k =6.06 x 10-2 sec -1
Similarly calculate for 50ºC
At 50º C, K 50 + 273 = 323 K
1/T = 1/323 = 3.1 x 10 -3 K-1
Plug the values we get \
ln k = -0.5
take antiln both side we get
k = 0.607 sec -1
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