Question

The rate constant for decomposition of ammonia on platinum surface is \(2.46 \times 10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}\). The rate of production of hydrogen is:

• A. \(1.23 \times 10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}\)
• B. \(4.92 \times 10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}\)
• C. \(7.38 \times 10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}\)
• D. \(14.76 \times 10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}\)

Hint:

For a reaction \[ aA \rightarrow bB \] always use \[ \text{Rate}=\frac{1}{b}\frac{d[B]}{dt} \] Thus, the rate of formation of a product is obtained by multiplying the reaction rate by its stoichiometric coefficient.

Answer 1

The Correct Option is C

Concept: For a reaction \[\begin{aligned} aA \rightarrow bB \end{aligned}\] the rate of reaction is related to the rate of formation of products as \[\begin{aligned} \text{Rate}=\frac{1}{b}\frac{d[B]}{dt} \end{aligned}\] For the decomposition of ammonia, \[\begin{aligned} 2NH_3 \rightarrow N_2 + 3H_2 \end{aligned}\] Hence, \[\begin{aligned} \text{Rate} =\frac{1}{3}\frac{d[H_2]}{dt} \end{aligned}\]

Step 1: Write the rate of reaction. Since the given rate constant has units of \(\mathrm{mol\,L^{-1}\,s^{-1}}\), the reaction is zero order and \[\begin{aligned} \text{Rate}=k =2.46\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned}\]

Step 2: Relate the rate of formation of hydrogen to the rate of reaction. From the stoichiometric equation, \[\begin{aligned} \text{Rate} =\frac{1}{3}\frac{d[H_2]}{dt} \end{aligned}\] Therefore, \[\begin{aligned} \frac{d[H_2]}{dt} &=3\times\text{Rate}\\ &=3\times2.46\times10^{-4}\\ &=7.38\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned}\]

Step 3: Tabulate the stoichiometric relationship. \[ \begin{aligned} \text{Rate constant, } k \quad & = \quad 2.46\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}} \\ \text{Stoichiometric coefficient of } H_2 \quad & = \quad 3 \\ \text{Rate of formation of } H_2 \quad & = \quad 3k \\ \text{Calculated value} \quad & = \quad 7.38\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned} \] \[\begin{aligned} \boxed{\frac{d[H_2]}{dt}=7.38\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}} \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.