Question

The following data were collected for the reaction between hydrogen and nitric oxide at \(700^{\circ} \mathrm{C}\) : $$ 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) $$ $$ \begin{array}{cccc} \text { Experiment } & {\left[\mathrm{H}_{2}\right](M)} & {[\mathrm{NO}](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 1 & 0.010 & 0.025 & 2.4 \times 10^{-6} \\ 2 & 0.0050 & 0.025 & 1.2 \times 10^{-6} \\ 3 & 0.010 & 0.0125 & 0.60 \times 10^{-6} \end{array} $$ (a) Determine the order of the reaction. (b) Calculate the rate constant. (c) Suggest a plausible mechanism that is consistent with the rate law. (Hint: Assume that the oxygen atom is the intermediate.)

Short answer

The reaction is first-order in \( \text{H}_2 \) and second-order in \( \text{NO} \), with a rate constant \( k = 0.384 \ \text{M}^{-2}\text{s}^{-1} \). A plausible mechanism involves formation and consumption of an OH intermediate.

Step-by-step solution

Write the General Rate Law

The rate law for the reaction can be generally expressed as \( \text{Rate} = k[\text{H}_2]^m[\text{NO}]^n \), where \(k\) is the rate constant, and \(m\) and \(n\) are the orders of the reaction with respect to \(\text{H}_2\) and \(\text{NO}\) respectively.

Determine the Order with Respect to H_2

Compare experiments 1 and 2. The concentration of \(\text{H}_2\) is halved while the \([\text{NO}])\) concentration remains constant. The rate is also halved. This indicates that the reaction is first-order with respect to \([\text{H}_2]\) (\(m = 1\)).

Determine the Order with Respect to NO

Compare experiments 1 and 3. Here, \([\text{NO}]\) is halved while \([\text{H}_2])\) remains constant. The rate is reduced by a factor of 4. This suggests a second-order dependence on \([\text{NO}]\) (\(n = 2\)).

Write the Specific Rate Law Equation

Based on the determined orders, the rate law becomes \( \text{Rate} = k[\text{H}_2]^{1}[\text{NO}]^2 \).

Calculate the Rate Constant, k

Using the rate law equation and data from experiment 1: \[2.4 \times 10^{-6} = k (0.010)^1(0.025)^2\]. Solving for \(k\), we find \(k = \frac{2.4 \times 10^{-6}}{(0.010)(0.000625)} = 0.384 \ \text{M}^{-2}\text{s}^{-1}\).

Propose a Mechanism

A plausible mechanism involves two steps. First, \([\text{H}_2]\) reacting with \([\text{NO}]\) to form an intermediate \(OH\), and then the intermediate reacting further. The simplest proposal \( \text{H}_2 + \text{NO} \rightarrow \text{OH} + \text{H} \) and \( \text{H} + \text{NO} \rightarrow \text{N} + \text{OH} \) fits these kinetics.

Key concepts

Reaction Order

Reaction order helps us understand how the concentration of reactants influences the rate of a chemical reaction. In the given reaction between hydrogen (\(\text{H}_2\)) and nitric oxide (\(\text{NO}\)), we need to determine how changes in concentrations affect the reaction rate. This is done through experiments.

For this reaction, by analyzing experiments 1 and 2, we see that halving the concentration of \(\text{H}_2\) results in halving the reaction rate. This shows a first-order dependence on \(\text{H}_2\), meaning the exponent associated with \(\text{H}_2\) in the rate law is 1.

• If a reactant's concentration changes and the rate changes proportionally, it's first-order for that reactant.
• If the rate changes by a factor of the concentration squared, it's second-order.

Similarly, when we compare experiments 1 and 3, halving the concentration of \(\text{NO}\) decreases the rate by a factor of four, indicating a second-order reaction with respect to \(\text{NO}\).

Rate Law

A rate law expresses the rate of a chemical reaction as a function of the concentration of the reactants. The general form is: \[\text{Rate} = k[\text{H}_2]^m[\text{NO}]^n\] where:

• \(k\) is the rate constant.
• \(m\) and \(n\) are the reaction orders determined experimentally for each reactant.

The specific rate law for this reaction was found to be: \[\text{Rate} = k[\text{H}_2][\text{NO}]^2\]due to the first-order dependence on \(\text{H}_2\) and second-order dependence on \(\text{NO}\). This means the reaction rate depends linearly on the concentration of hydrogen and quadratically on the concentration of nitric oxide.

Understanding the rate law is key to predicting how the exchange in reactant concentration will influence the reaction rate.

Rate Constant

The rate constant, denoted as \(k\), is a crucial component of the rate law. The value of \(k\) is independent of the reactant concentrations but is highly sensitive to temperature. It's calculated using experimental data.

From the solved example, using data from experiment 1, the rate constant \(k\) was determined to be 0.384 \(\text{M}^{-2}\text{s}^{-1}\). The calculation used is straightforward:

• Insert known rate and concentrations into the rate law.
• Rearrange to solve for \(k\).

This constant helps us understand the intrinsic rate at which a reaction proceeds under defined conditions, excluding concentration effects.

Reaction Mechanism

A reaction mechanism proposes the step-by-step sequence of elementary reactions by which the overall chemical change occurs. For the hydrogen and nitric oxide reaction, the proposed mechanism included intermediate species.

The hint suggests an oxygen atom as an intermediate, leading to a plausible mechanism:

• First step (\(\text{slow}\)): \(\text{H}_2 + \text{NO} \rightarrow \text{OH} + \text{H}\)
• Second step: \(\text{H} + \text{NO} \rightarrow \text{N} + \text{OH}\)

This mechanism is consistent with observed kinetics, involving a complex balance of formation and consumption of intermediates. It's vital to suggest mechanisms that align with the experimentally determined rate law and respect the molecularity of each proposed elementary step, like bimolecular reactions.