Question

For first order decomposition,
X- Y+Z. rate constant is given by
k =(4.6 x 1015 e 54000/ KT
where, the energy of activation is in
calories. Calculate
temperature at which X would
decompose at the rate of 1% per second,
(ii) temperature where decomposition is
80% complete in ih.

Answer 1

The temperature is K = 434.3°C.

Explanation:

(i) It is given that:

k = 4.6×10^14  e^−54000 RT/     -------(1)

We know that:

K = 2.303 / t log a / a−x  

K = 2.3031 log 100 / 100−1

K = 2.303 log 100 / 99

K = 2.303 log(1.010)

K  =0.0099 s−1

Substituting the value of k in (i)

0.0099 = 4.6 × 10^14 e^−54000 / RT

OR log 0.0099  = log 4.6 × 10^14  540002.303×1.987×T

−2.0 = 13.337 − 11800.7 T

T = 11800.7 / 15.337

T = 769.42 K = 496.42°C

(ii) k2 = 2.303 / 60×60 log 100 / 100 − 80

= 2.3033 / 600 log 5  = 0.00045 s−1

Substituting the value of k2 in (1)

0.00045= 4.6 × 10^14 e^−54000/RT or, log 0.00045

= log 4.6 × 10^14 − 54000 / 2.303 × 1.987 × T

−3.347 = 13.337 − 11800.7 T

or, T = 11800.7 / 16.684

T = 707.3 K = 434.3°C

Thus the temperature is K = 434.3°C.