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Q3.3:
The decomposition of \(NH_{3}\) on platinum surface is zero order reaction. What are the rates of production of \(N_{2}\) and \(H_{2}\) if \(k = 2.5 \times 10^{-4} mol^{-1} L s^{-1}\)?

Solution :

Step 1: Establishing the Chemical Equation and Rate Law

The decomposition of ammonia ($NH_{3}$) on a hot platinum surface is a zero-order reaction. The balanced chemical equation is:

$2NH_{3}(g) \xrightarrow{Pt} N_{2}(g) + 3H_{2}(g)$

For a reaction $aA \rightarrow bB + cC$, the rate of reaction is defined as [Per the Law of Mass Action and Definition of Reaction Rate]:

$\text{Rate} = -\frac{1}{a} \frac{d[A]}{dt} = \frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt}$

For the given reaction, the rate expression is:

$\text{Rate} = -\frac{1}{2} \frac{d[NH_{3}]}{dt} = \frac{d[N_{2}]}{dt} = \frac{1}{3} \frac{d[H_{2}]}{dt}$

Step 2: Identifying Given Values

The reaction is zero order, meaning the rate of the reaction is independent of the concentration of the reactant. Thus, the rate of the reaction is equal to the rate constant $k$.

  • $k = 2.5 \times 10^{-4} \, mol \, L^{-1} s^{-1}$
  • $\text{Rate of Reaction} = k = 2.5 \times 10^{-4} \, mol \, L^{-1} s^{-1}$

Step 3: Calculating Rate of Production of $N_{2}$

From the stoichiometric relationship established in Step 1:

$\text{Rate of production of } N_{2} = \frac{d[N_{2}]}{dt} = \text{Rate of Reaction}$

$\frac{d[N_{2}]}{dt} = 2.5 \times 10^{-4} \, mol \, L^{-1} s^{-1}$

Step 4: Calculating Rate of Production of $H_{2}$

From the stoichiometric relationship:

$\frac{d[H_{2}]}{dt} = 3 \times \text{Rate of Reaction}$

Substituting the value of $k$:

$\frac{d[H_{2}]}{dt} = 3 \times (2.5 \times 10^{-4} \, mol \, L^{-1} s^{-1})$

$\frac{d[H_{2}]}{dt} = 7.5 \times 10^{-4} \, mol \, L^{-1} s^{-1}$

Visualizing the Reaction Dynamics

Reaction Progress (Time) Concentration [H2] (3k) [N2] (k)

Summary of Rates

Product Rate Expression Calculated Rate ($mol \, L^{-1} s^{-1}$)
$N_{2}$ $d[N_{2}]/dt$ $2.5 \times 10^{-4}$
$H_{2}$ $d[H_{2}]/dt$ $7.5 \times 10^{-4}$

Final Solution: The rate of production of $N_{2}$ is $2.5 \times 10^{-4} \, mol \, L^{-1} s^{-1}$ and the rate of production of $H_{2}$ is $7.5 \times 10^{-4} \, mol \, L^{-1} s^{-1}$.


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