## R

reactants at the beginning of the experiment, and at and bt are their concentrations at time t. The value of the final term of equation (4.70) rapidly tends to zero as kt becomes greater than k0. Thus a graph of log f (c) against log(1 + t) will be linear from that time after which k0 is negligible in comparison with kt. The slope of the line is (1 + Eab/R), enabling Ea to be determined. The rate constant K

Substituting equation (4.65) into equation (4.66) gives

For first-order reactions -dc/dt = kc, where c is concentration. Substituting for k from equation (4.68) and integrating gives then be calculated from the intercept when log(1 + t) = 0, which is equal to log k0 -log(1 + Eab/R). The rate constant at any other temperature may be calculated from k0 and Ea.

EXAMPLE 4.10 Accelerated storage testing using temperature-time programme

In a study of the first-order decomposition of riboflavin in 0.05 mol dm 3 NaOH using accelerated storage techniques, the temperature was programmed to rise from 12.5 to 55°C using a programme constant, b, of 2.171 x 104 K_1. The initial concentration, c0, of riboflavin was 10 4 mol dm 3, and the concentration ct remaining at time t was as follows:

0.585 0.996 1.512 2.163 2.982 4.013 5.312 6.946 5c, 9.881 9.763 9.532 9.109 8.371 6.902 4.931 2.435 (mol dm03)

where c0 and ct are the concentrations at zero time and at time t, respectively. Therefore,

Eab\ / Eab log f (c) = log k0 - log 1 + -R-1 + j 1 +

Calculate the activation energy and the rate constant at 20°C.

Answer

For first-order reactions the data are plotted according to equations (4.70) and (4.71).

A similar equation applies to second-order reactions with

, „ 2.303 at 2.303 b0 log f (c) =-- log -t +-- log a0 - b0 bt a0 - b0 a0

where a0 and b0 are the concentrations of the

t |
log(1 +1 ) |
log[2.303 log(c0/cf)j |

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