The pressure of a gas decomposing at the surface of a solid catalysts has been measured at different times and the results are given below:
t(s) 0 100 200 300
p (Pa) $4.00 \times 10^{- 3}$
$3.50 \times 10^{- 3}$ $3.00 \times 10^{- 3}$ $2.5 \times 10^{- 3}$
The half life period of the reaction is(in sec)

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Solution

(a) It can be seen that the rate of reaction between different time intervals is :
0 - 100 s, rate = $\frac{[4 - 3.5] 0^{3} P a}{100} = 5 P a s^{- 1}$
100 - 200 s, rate = $\frac{[3.50 -, 3.00] \times 10^{3} P a}{100 s} = 5 P a s^{- 1}$
200 - 300 s, rate = $\frac{[3.00 - 2.50] \times 10^{3} P a}{100} = 5 P a s^{- 1}$
We notice that the rate remains constant and therefore, the reaction is of zero order. Alternatively, if we plot p againts t, it is a straight line again indicating it is a zero order reaction.
(b) k = Rate = 5 Pa $s^{- 1}$
(c) $t_{1 / 2} = \frac{i n i t i a l c o n c e n t r a t i o n o r p r e s s u r e}{2 k}$
= $\frac{4.00 \times 10^{3} P a}{2 \times 5 P a S^{- 1}} = 400 s$

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t(s)	0	100	200	300
p (Pa)	$4.00 \times 10^{- 3}$	$3.50 \times 10^{- 3}$	$3.00 \times 10^{- 3}$	$2.5 \times 10^{- 3}$

The pressure of a gas decomposing at the surface of a solid catalysts has been measured at different times and the results are given below:t(s)0100200300p (Pa)4.00×10−33.50×10−33.00×10−32.5×10−3The half life period of the reaction is(in sec)