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The rate constant for the decomposition of N_(2)O_(5) at various temperatures is given below: Draw a graph beetween In k and 1//T and calculate the values of A and E_(a). Predict the rate constant at 30^(@) and 50^(@)C |
Answer» Solution :To draw the plot of log K versus `1//T`, we can re-write the given DATA as follows:  Draw the graph as shown on the next page, from the graph, we find that Slope = `-2.4/(0.00074) = -E_(a)/(2.303 R)` `therefore` Activation energy, `(E_(a)) = (2.4 xx 2.303 xx R)/(0.00047) = (2.4 xx 2.303 xx 8.314 J mol^(-1))/(0.00047)` `=17, 689 J mol^(-1)= 17.689 kJ mol^(-1)` As we know log k = log a -`(E_(a))/(2.303 RT)` (Compare it with y = mx+c which is equation of line in intercept form) `log k = (-E_(a))/(2.303RT)(1/T) + log A` log A = VALUE of intercept on y-axis i.e., on the k-axis. `=(-1+7.2) = 6.2``[y_(2)-y_(1) = -1-(-7.2)]` Frequency factor, A= ANTILOG `6.2 = 1585000 = 1.585 xx 10^(6)` collisions `s^(-1)` The value of rate constant k can be found from the graph as follows:
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