Question

1. Consider the following reaction: 2 ClO2 (aq) + 2 I- (aq)  2 ClO2 - (aq) + I2 (aq) The order with respect to ClO2 was determined by starting with a large excess of I- , so that it its concentration was effectively constant. This simplifies the rate law to: rate = k’[ClO2] m where k’ = k[I- ] n . Determine the order with respect to ClO2 and the rate constant k’ by plotting the following data assuming first-order, then second-order kinetics. You need not attach the plots for these two treatments, but you should report the linear regression equations and correlation coefficient for each. (Sample Exercise 13.4) (7 pts)

Time (s) [ClO2] (moles/liter) 0.00 0.000477, 1.00 0.000431 ,2.00 0.000391, 3.00 0.000353 ,5.00 0.000289, 10.00 0.000176 ,30.00 0.000024, 50.00 0.0000032

2. The reaction, 2 NOCl (g)  2 NO (g) + Cl2 (g), has rate-constant values of 9.3 x 10-6 s -1 at 350 K and 6.9 x 10-4 s -1 at 400 K. Calculate the activation energy for the reaction and the rate constant at 425 K (

Answer 1

Rate = K' [ClO2]^m, m is the order of reaction. For solving this problem, integral method is adopted.

-dCl2O2/dt= K'[ ClO2]^m

when m= 1it is 1st order kinetics. -d(CLO2)/dt= K'[ClO2], when integrated, ln (ClO2)= ln(ClO2)o-K't, where (ClO2)o is the concentration of ClO2 at zero time, (ClO2) is concentration is concentration of ClO2 at any time. So a plot of ln (ClO2) vs time gives straight line whose slope is -K' if 1st order kinetics is obeyed.

when m=2, it is second order reaction, -dClO2/dt= K(ClO2)2, when integrated, 1/(ClO2)= 1/(ClO2)0+Kt

So a plot of 1/CoO2 vs time gives straight line if second order kinetics is obeyed. both plots are shown below

1st order plot

2nd order plot

the plot suggests 1st order kinetics ( since the plot of lnClO2 vs time is straight line

the slope is = -0.1/sec, K'= Rate constant= 0.1/sec

Hence the rate equation becomes -dClO2/dt= 0.1(ClO2)

2. Arhenius equation can be used to calculate the activation enrrgy of the reaction since the rate constant (K) at two different temperatures are given

Arhenius equartion at two different temperatures can be written as

ln(K2/K1)= (E/R)*(1/T1-1/T2)

K2= rate constant at 400K= 6.9*10-4/sec T2= 400K, K1= rate constant at 350K= 9.3*10-6/sec, E= activation energy and R= gas constant =8.314 J/mole.K

ln( 6.9*10^-4/ 9.3*10^-6)= (E/8.314)*(1/350-1/400), E= 100256 J/mole

At 425K, K2 needs to be determined assuminig K1= 6.9*10-4 at T=400K

ln(K2/6.9*10-4)= (100256/8.314)*(1/400-1/425) =1.77

K2/6.9*10^-4= 5.87, K2= 0.004051/sec