Question

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

t (sec)	P(mm of Hg)
0	35.0
360	54.0
720	63.0

Calculate the rate constant.

Answer 1

The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

                             \((CH_3)_2CHN=NCH(CH_3)_2(g)→N_2(g)+C_6H_{14}(g)\)
At t = 0                                            P₀                                    0                0
At t = 0                                         P₀ - p                                  p               p
P_t = (P₀ - p) + p + p
After time t, total pressure
⇒ P_t = P₀ + p
⇒ p = P_t - P₀
Therefore, P₀ - P = P₀ - (P_t - P₀)
= 2P₀ - P_t
For a first order reaction, 

\(k = \frac {2.303}{t} log \ \frac {P_0}{P_0-p}\)

\(k = \frac {2.303}{t} log\ \frac {P_0}{2P_0-p_t}\)

\(When\  t = 360\  s\)

\(k = \frac {2.303}{360\ s} log\ \frac {35.0}{2\times 35.0-5.0}\)

\(k = 2.175 \times 10^{-3} s^{-1}\)

\(When\  t = 720\  s\)

\(k = \frac {2.303}{720\ s} log\ \frac {35.0}{2\times 35.0-63.0}\)

\(k = 2.235 \times 10^{-3} s^{-1}\)

Hence the average value of rate constant is 

\(k =\frac { (2.175 \times 10^{-3}) + (2.235 \times 10^{-3})}{2} s^{-1}\)

\(k = 2.21\times 10^{-3} s^{-1}\)