Half Life - Test Method

The thermal decomposition of nitrous oxide (N₂O) in the gas phase at 1030 K is studied in a constant volume vessel at various initial pressures of N₂O. The half-life data so obtained are as follows:

po (mm Hg)	52.5	139	290	360

t1/2 (sec)	860	470	255	212

Determine the rate equation that fits the data.

Calculations:

t_1/2 Vs. C_Ao in a log-log plot is a straight line with a slope of 1-n

C_Ao = p_Ao/RT = p_o/RT (for an ideal gas)

ln(C_Ao) Vs. ln(t_1/2) in a linear XY plot gives a straight line, and the slope of the line is 1 - n

from the above figure,

therefore, n = 1 + 0.7469 = 1.7469 = 1.75

i.e., Order of the reaction (n) = 1.75

t_1/2 and C_Ao are related by the formula:

${\displaystyle {�}_{1/2}=\frac{2^{�-1}-1}{�(�-1)}{�}_{��}^{1-�}}$

for a value of p_o = 290, and t_1/2 = 255,

C_Ao = (290/760) x 1.01325 x 10^-5/(8314 x 1030) = 4.515 x 10^-3 gmol/lit.

therefore,

k = (2^{1.75 - 1} - 1) x (4.515 x 10^-3)^(1-1.75) / (255 x (1.75 - 1)) = 0.2046 sec^-1.(gmol/lit) ^-0.75

the rate expression for the given reaction is,

-r_N2O = 0.2046 (C _N2O)^1.75