Question

in investigation is performed in which the concentration of nitrogen dioxide reacting is measured as a function of time. Time (S) [NO2](mol/L) 0.500 0.445 0.380 0.340 0.250 0.175 12 30 90 180 a) Plot a graph of (NO) vs. time (5 marks) b) Determine the average rate of reaction of NO2 between 10 and 60s (2 marks) c) As the [NO2) halved, what was the effect on the instantaneous rate of reaction? (2 marks)

Answer 1

Graph is as follows: For zero order reaction the graph [A] versust is linear For first order reaction the graph In (A) versust is linear. For Second order reaction the graph versus t is linear. This is an example of second order. When the [NO2] halved then the instantaneous rate of reaction is decreased by four times 0+ [NO2] time → 1.NO.1 form the graph first calculate the [NO2] at 10 s and 60 s which are as follows: [NO2]at 10s=0.4536 M [NO2 lat 60 =0.3052 M time → Now average rate= ([NO2]at 60 s)-([NO2]at 10 s) 60-10 = 0.3052-0.4536/50 = -2.968*10^-3