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Chemical Kinetics

Question
CBSEENCH12006398

The rates of a reaction starting with initial concentration 2 x 10–3 M and 1 x 10–3M are equal to 2.40 x 10–4 M s–1 and 0.60 x 10–4 M s–1 respectively. Calculate the order of the reaction with respect to the reactant and also the rate constant.

Solution
r0 = kR0n(r0)1(r0)2/ = kR01R02n              a  = logr01/r02logR01/R02                   = log(2.40 × 10-4/0.60 × 10-4log (2 × 10-3/1×10-3)                   = log 4log 2=2

Thus reaction is of second order. The rate constant

k=rateA2   = 2.4 ×10-4 Ms-1/(2 × 10-3 M)2   = 0.6 × 102 mol-1 Ls-1.

Use of integrated Rate Equations:
This method is also known as the method of trial and error. The kinetic data are fitted to different integrated rate equations. Whenever if the data fits with the equation for the correct order of the reaction, it will give constant value of rate constant for all data points (concentrations at different times). These equations also lead to straight lines when appropriate function of the concentration is plotted against time ‘t’. For example, for zero reaction, a plot between concentration and time gives a straight line with slope of the line equal to k. Similarly, for the first order reaction, a graph between ln (R) against t gives a straight line with slope equal to – k.