The decomposition of NH3 on platinum surface is zero order reaction. If k = 2.5 × 10-4 mol L-1 s-1 the rate of production of H2 is
Understanding Zero Order Reactions and Decomposition of Ammonia
The question describes the decomposition of ammonia (\(\text{NH}_3\)) on a platinum surface as a zero-order reaction. This means the rate of the reaction is independent of the concentration of the reactant (\(\text{NH}_3\)). The rate constant (\(k\)) for this reaction is given as \(2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{ s}^{-1}\). We need to determine the rate of production of hydrogen (\(\text{H}_2\)).
Balanced Chemical Equation
The decomposition of ammonia on a platinum catalyst is represented by the following balanced chemical equation:
\[ 2\text{NH}_3\text{(g)} \xrightarrow{\text{Pt catalyst}} \text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \]Relating Reaction Rate to Species Rates
For a general reaction \(aA \rightarrow bB + cC\), the rate of the reaction is defined as:
\[ \text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} \]where \(-\frac{d[A]}{dt}\) is the rate of disappearance of reactant A, and \(\frac{d[B]}{dt}\) and \(\frac{d[C]}{dt}\) are the rates of appearance of products B and C, respectively. The coefficients \(a\), \(b\), and \(c\) are the stoichiometric coefficients from the balanced equation.
Applying this to the decomposition of ammonia:
\[ \text{Rate} = -\frac{1}{2}\frac{d[\text{NH}_3]}{dt} = +\frac{1}{1}\frac{d[\text{N}_2]}{dt} = +\frac{1}{3}\frac{d[\text{H}_2]}{dt} \]Zero Order Kinetics and Rate Constant
The problem states that the reaction is zero order with respect to ammonia on the platinum surface. For a zero-order reaction, the rate of the reaction is equal to the rate constant \(k\), and it does not depend on the concentration of the reactant.
\[ \text{Rate} = k \]Given \(k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{ s}^{-1}\).
Calculating Rate of Production of H\(_2\)
We want to find the rate of production of \(\text{H}_2\), which is \(\frac{d[\text{H}_2]}{dt}\). From the rate expression, we have the relationship:
\[ \text{Rate} = +\frac{1}{3}\frac{d[\text{H}_2]}{dt} \]Since the Rate is equal to \(k\) for a zero-order reaction:
\[ k = \frac{1}{3}\frac{d[\text{H}_2]}{dt} \]To find the rate of production of \(\text{H}_2\), we rearrange the equation:
\[ \frac{d[\text{H}_2]}{dt} = 3 \times k \]Substitution and Final Calculation
Substitute the given value of \(k\) into the equation:
\[ \frac{d[\text{H}_2]}{dt} = 3 \times (2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{ s}^{-1}) \] \[ \frac{d[\text{H}_2]}{dt} = 7.5 \times 10^{-4} \, \text{mol L}^{-1} \text{ s}^{-1} \]Thus, the rate of production of \(\text{H}_2\) is \(7.5 \times 10^{-4} \, \text{mol L}^{-1} \text{ s}^{-1}\).
Let's look at the given options:
- Option 1: \(2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{ s}^{-1}\)
- Option 2: \(7.5 \times 10^{-4} \, \text{mol L}^{-1} \text{ s}^{-1}\)
- Option 3: \(5.0 \times 10^{-4} \, \text{mol L}^{-1} \text{ s}^{-1}\)
- Option 4: \(10.0 \times 10^{-4} \, \text{mol L}^{-1} \text{ s}^{-1}\)
Our calculated rate of production of \(\text{H}_2\) matches Option 2.
Key Concepts in Reaction Kinetics
This problem involves several key concepts from chemical kinetics:
- Reaction Rate: How fast reactants are consumed or products are formed.
- Rate Law: An equation that relates the reaction rate to the concentrations of reactants. For a zero-order reaction, the rate law is Rate = \(k\).
- Order of Reaction: The sum of the exponents of the concentration terms in the rate law. A zero-order reaction has an order of 0.
- Rate Constant (\(k\)): A proportionality constant in the rate law that relates the rate of reaction to the concentrations of reactants. It is temperature-dependent.
- Stoichiometry: The quantitative relationship between reactants and products in a balanced chemical equation. Stoichiometric coefficients are used to relate the rate of the reaction to the rates of change of individual species.
| Term | Definition | Relevance to Problem |
|---|---|---|
| Zero Order Reaction | Reaction rate is independent of reactant concentration. | Given property of NH3 decomposition on Pt. |
| Rate Constant (k) | Proportionality constant in the rate law. | Given value used to calculate product rate. |
| Rate of Production | Rate at which a product's concentration increases over time. | The quantity we needed to calculate for H2. |
| Stoichiometry | Coefficients in balanced equation. | Used to relate overall reaction rate to the rate of H2 formation (factor of 3). |
Revision Table: Chemical Kinetics Basics
| Order | Rate Law Example | Rate Unit | k Unit | Integrated Rate Law Example |
|---|---|---|---|---|
| Zero Order | Rate = k | mol L-1 s-1 | mol L-1 s-1 | [A]t = [A]0 - kt |
| First Order | Rate = k[A] | mol L-1 s-1 | s-1 | ln[A]t = ln[A]0 - kt |
| Second Order | Rate = k[A]2 or Rate = k[A][B] | mol L-1 s-1 | L mol-1 s-1 | 1/[A]t = 1/[A]0 + kt |
Additional Information: Surface Reactions and Zero Order
The decomposition of ammonia on a platinum surface is an example of a heterogeneous catalytic reaction. In such reactions, the reactants are adsorbed onto the surface of the catalyst (platinum in this case). The reaction then occurs on the surface. Zero-order kinetics are often observed in heterogeneous catalysis when the reactant is strongly adsorbed onto the catalyst surface.
When the reactant is strongly adsorbed, the surface sites become saturated with reactant molecules. This means that almost all available surface sites are occupied by \(\text{NH}_3\) molecules. As more \(\text{NH}_3\) molecules approach the surface, they cannot find empty sites to adsorb onto. Therefore, increasing the concentration of \(\text{NH}_3\) in the gas phase does not lead to a significant increase in the number of adsorbed \(\text{NH}_3\) molecules on the surface, where the reaction actually happens.
Since the rate of the surface reaction depends on the amount of adsorbed reactant, and this amount is essentially constant (at saturation) regardless of the gas phase concentration, the overall reaction rate becomes independent of the gas phase \(\text{NH}_3\) concentration. This results in zero-order kinetics.
The rate constant \(k\) for such a process includes factors like the intrinsic rate constant of the surface reaction and the surface area of the catalyst.
Important Questions from Kinetics of Reaction
The molecularity of the following elementary reaction is NH4NO2 → N2 + 2H2O
