The thermal decomposition of nitrous oxide (N_{2}O) in the gas phase at 1030^{o}K is studied in a constant volume vessel at various initial pressures of N_{2}O. The half-life data so obtained are as follows:

p_{o}(mm Hg) 52.5 139 290 360 t_{1/2}(sec) 860 470 255 212

2N_{2}O à
2N_{2} + O_{2}

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,

1-n = -0.7469

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:

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}