Problem 26 - Chemistry The Molecular Science

Question

For the reaction \(2 \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) these data were obtained at \(1100 \mathrm{~K}\) : \begin{tabular}{ccc} \hline [NO] (mol/L) & {\(\left[\mathrm{H}_{2}\right](\mathrm{mol} / \mathrm{L})\)} & Initial Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline \(5.00 \times 10^{-3}\) & \(2.50 \times 10^{-3}\) & \(3.0 \times 10^{-3}\) \\\ \(15.0 \times 10^{-3}\) & \(2.50 \times 10^{-3}\) & \(9.0 \times 10^{-3}\) \\ \(15.0 \times 10^{-3}\) & \(10.0 \times 10^{-3}\) & \(3.6 \times 10^{-2}\) \\ \hline \end{tabular}

Step-by-Step Solution

Answer

The rate law is \( \text{Rate} = k[NO][H_2] \) with \( k = 240 \text{ L}^2 \text{mol}^{-2} \text{s}^{-1} \).

Step 1: Determine the Rate Law Form

For a general reaction, the rate law can be expressed as:\[ \text{Rate} = k [NO]^m [H_2]^n \]where \( k \) is the rate constant, and \( m \) and \( n \) are the reaction orders with respect to \( NO \) and \( H_2 \), respectively. We need to determine \( m \) and \( n \) using the provided data.

Step 2: Calculate Reaction Order for [NO]

Compare Experiments 1 and 2 where \([H_2]\) remains constant at \(2.50 \times 10^{-3} \text{ mol/L}\) but \([NO]\) changes from \(5.00 \times 10^{-3}\) to \(15.0 \times 10^{-3}\):1. The rate increases from \(3.0 \times 10^{-3}\) to \(9.0 \times 10^{-3} \text{ mol L}^{-1} \text{s}^{-1}\).2. As \([NO]\) triples, the rate also triples.This means the reaction is first order in \([NO]\), i.e., \(m=1\).

Step 3: Calculate Reaction Order for [H2]

Compare Experiments 2 and 3 where \([NO]\) remains constant at \(15.0 \times 10^{-3} \text{ mol/L}\) but \([H_2]\) changes from \(2.50 \times 10^{-3}\) to \(10.0 \times 10^{-3}\):1. The rate increases from \(9.0 \times 10^{-3}\) to \(3.6 \times 10^{-2} \text{ mol L}^{-1} \text{s}^{-1}\).2. As \([H_2]\) quadruples, the rate increases fourfold.This means the reaction is first order in \([H_2]\), i.e., \(n=1\).

Step 4: Write the Rate Law

Based on our calculations:\[ \text{Rate} = k [NO]^1 [H_2]^1 = k [NO][H_2] \]

Step 5: Determine the Rate Constant k

Utilize any of the experimental data points to solve for \( k \). Using the first experiment (\([NO]=5.00 \times 10^{-3} \text{ mol/L}, [H_2]=2.50 \times 10^{-3} \text{ mol/L}\)):\[ 3.0 \times 10^{-3} = k \times (5.00 \times 10^{-3}) \times (2.50 \times 10^{-3}) \]\[ k = \frac{3.0 \times 10^{-3}}{12.5 \times 10^{-6}} = 240 \text{ L}^2 \text{mol}^{-2} \text{s}^{-1} \]

Key Concepts

Rate Law

When studying reaction kinetics, the rate law is a fundamental equation that helps us understand how the concentrations of reactants affect the reaction rate. It is expressed as:

• \[ \text{Rate} = k [A]^m [B]^n \] where [A] and [B] are the concentrations of the reactants, and \(m\) and \(n\) are the reaction orders corresponding to each reactant.
• The rate constant \(k\) is a proportionality factor that remains constant as long as the temperature is constant.

The rate law is empirically determined. This means it is based on experimental data rather than derived from the chemical equation of the reaction. It tells us how sensitive the rate is to changes in the concentration of each reactant, which is crucial for predicting and controlling the speed of chemical reactions.

Reaction Order

The reaction order in kinetics refers to the power to which the concentration of a reactant is raised in the rate law. It's vital for understanding how variations in reactant concentrations influence the rate at which a reaction proceeds. Here's a simple breakdown:

• If a reaction is first order with respect to a particular reactant, it means that doubling the concentration of that reactant will double the reaction rate.
• In the provided exercise, comparing experiments shows that the reaction is first order concerning both \([NO]\) and \([H_2]\).
• Reaction orders can be whole numbers, but they may also be fractions or negative values, depending on the specific reaction.

Understanding reaction orders is integral to controlling reaction outcomes, especially in industrial and laboratory settings where optimizing efficiency and product yield is key.

Rate Constant

The rate constant, \(k\), plays a crucial role in the rate law, linking the reaction rate with the concentrations of reactants raised to their respective reaction orders. This constant is specific to a given reaction and depends on several factors:

• Temperature: The rate constant generally increases with temperature due to the increased kinetic energy of molecules, making collisions more frequent and energetic.
• Catalysts: These substances can alter \(k\) by providing an alternative reaction pathway with a lower activation energy.
• Units: The units of \(k\) vary depending on the reaction order. In the exercise, \(k\) has units of L²mol⁻²s⁻¹, reflecting a second-order overall reaction.

Finding \(k\) accurately is essential as it allows for the prediction of reaction rates under varying conditions of concentration and temperature. This helps in tailoring processes for desirable speeds and outcomes in both theoretical and applied chemistry.