Half lives of decomposition of $N H_{3}$ on a catalyst surface for difference initial pressure(P) are given as
P(atm) 200 300 500
$t_{\frac{1}{2}}$ (min) 10 15 25
Then, the order of the reaction is________.

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Solution

The correct option is C 0
$\frac{t_{\frac{1}{2}} (1)}{t_{\frac{1}{2}} (2)} = (\frac{a_{2}}{a_{1}})^{n - 1}$

$\frac{10}{15} = (\frac{300}{200})^{n - 1}; (\frac{2}{3})^{1} = (\frac{3}{2})^{n - 1}$ $= (\frac{2}{3})^{1 - n} \Rightarrow n = 0$

Now, for zero-order reaction:

$t_{1 / 2} = \frac{P}{2 k}$
$k_{1} = \frac{P}{2 t_{1 / 2}} = \frac{200}{20} = 10$
$k_{2} = \frac{300}{2 \times 15} = \frac{300}{30} = 10$
$k_{3} = \frac{500}{2 \times 25} = \frac{500}{50} = 10$
$S o, k_{1} = k_{2} = k_{3}$
So, it is zero reaction

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Half lives of decomposition of NH3 on a catalyst surface for difference initial pressure(P) are given asP(atm) 200 300 500t12(min) 10 15 25Then, the order of the reaction is________.

The correct option is C 0t12(1)t12(2)=(a2a1)n−11015=(300200)n−1;(23)1=(32)n−1 =(23)1−n⇒n=0Now, for zero-order reaction:t1/2=P2kk1=P2t1/2=20020=10k2=3002×15=30030=10k3=5002×25=50050=10So, k1=k2=k3So, it is zero reaction

Half lives of decomposition of $N H_{3}$ on a catalyst surface for difference initial pressure(P) are given as
P(atm) 200 300 500
$t_{\frac{1}{2}}$ (min) 10 15 25
Then, the order of the reaction is________.