Question

The decomposition of $N{H_3}$ on the platinum surface is a zero order reaction. What are the rates of production of ${N_2}$ and ${H_2}$ if $k = 2.5 \times {10^{ - 4}}mo{l^{ - 1}}L{s^{ - 1}}$ ?

Answer 1

In order to this question, to find the rate of production of ${N_2}$ and ${H_2}$ from the given decomposition reaction, we will first rewrite the balanced reaction as per the question and then we will apply the zero order reaction formula for the reaction. And finally we will apply the formula to find the rate of production for the given elements of the reaction, $\because Rate\,of\,production\,of\,element = No.\,of\,moles\,of\,element \times k$ .

Complete answer: According to the question, the given decomposition reaction is:
$2N{H_3}\mathop \to \limits^{Pt} {N_2} + 3{H_2}$
Now, the rate of the above reaction is: (apply the zero order reaction)
$Rate = - \dfrac{1}{2}[\dfrac{{d[N{H_3}]}}{{dt}}] = [\dfrac{{d[{N_2}]}}{{dt}}] = + \dfrac{1}{3}[\dfrac{{d[{H_2}]}}{{dt}}]$
As the order of the reaction is zero, so the rate of reaction-
$Rate\,of\,reaction = k$
Given that:
$k = 2.5 \times {10^{ - 4}}mo{l^{ - 1}}L{s^{ - 1}}$
$\because Rate\,of\,production\,of\,{N_2} = No.\,of\,moles\,of\,{N_2} \times k$
$Rate\,of\,production\,of\,{N_2} = 1 \times 2.5 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}} = 2.5 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}}$
Similarly,
$\because Rate\,of\,production\,of\,{H_2} = No.\,of\,moles\,of\,{H_2} \times k$
$Rate\,of\,production\,of\,{H_2} = 3 \times 2.5 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}} = 7.5 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}}$

Note:
A decomposition process occurs when chemical bonds are broken. The use of heat to initiate a breakdown reaction is common. Decomposition reactions can be used to analyse materials by scientists.