From the following data for the decomposition of azoisopropane at 543 K into hexane and nitrogen, calculate the average value of rate constant and also show that it is a first order reaction.

Explanation

Azoisopropane dissociates into hexane and nitrogen as follows:
` underset("Azoisopropane")((CH_3)_2 CHN = NCH (CH_3)_2) overset(" 543 K") to underset("Hexane ")(C_6 H_(14) ) + N_2`
Initial pressure of azoisopropane, `p_i` = 35.00
Since for every molecule of azoisopropane that decomposes, two gaseous molecules are produced, the pressure increases with passage of time.
Total pressure, `P_t = P_A + PN_2 +PC_6 H_(14)`
If at any instant of time, x is the decrease in pressure of azoisopropane, then increase in pressure each of `N_2` and `C_6 H_(14)` would be x.
` p_A = p_i - x , P_(N_2) = x , P_(c_6 H_(14)) = x `
` therefore p_t = (p_i - x) +x +x=p_i + x `
` x=p_t -p_i`
` p_A = (p_i-x) =p_i -(p_t -p_i ) = 2p_i -p_t`
If the reaction follows first order kinetics, it must obey the equation
` k=( 2.303 )/(t) log"" (a)/((a-x)) = ( 2.303 )/(t) log"" (p_i)/( 2p_i - p_t)`
In the present case, `P_i = 35` so that `2p_i` = 70.00 The value of k at different instants can be calculated as follows:

Since the value of k is almost constant, the reaction is a first order reaction.
Average value of rate constant is `2.205 xx 10^(-3) s^(-1)`