Problem 12 - Chemistry and Chemical Reactivity

Question

The reaction $$2 \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ was studied at \(904^{\circ} \mathrm{C},\) and the data in the table were collected. $$\begin{array}{lll}\hline \begin{array}{l}\text { Reactant Concentration } \\\\(\mathrm{mol} / \mathrm{L})\end{array} & & \\\\\hline[\mathrm{N} 0] & {\left[\mathrm{H}_{2}\right]} & \begin{array}{l}\text { Rate of Appearance of } \mathrm{N}_{2} \\\\(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s})\end{array} \\\\\hline 0.420 & 0.122 & 0.136 \\\0.210 & 0.122 & 0.0339 \\\0.210 & 0.244 & 0.0678 \\\0.105 & 0.488 & 0.0339 \\\\\hline\end{array}$$ (a) Determine the order of the reaction for each reactant. (b) Write the rate equation for the reaction. (c) Calculate the rate constant for the reaction. (d) Find the rate of appearance of \(\mathrm{N}_{2}\) at the instant when \([\mathrm{NO}]=0.350 \mathrm{mol} / \mathrm{L}\) and \(\left[\mathrm{H}_{2}\right]=0.205 \mathrm{mol} / \mathrm{L}.\)

Step-by-Step Solution

Answer

(a) Order: NO = 2, H2 = 1; (b) Rate = k [NO]^2 [H_2]; (c) k ≈ 6.18 L^2/mol^2·s; (d) Rate ≈ 0.154 mol/L·s.

Step 1: Analyze Given Data for Reactant Orders

To determine the order of each reactant, compare trials where the concentration of one reactant changes while the other stays constant. Start with NO. Compare trials 1 and 2: \[ \left( \frac{0.420}{0.210} \right)^m = \frac{0.136}{0.0339} \] Solve for \( m \): \[ 2^m = 4 \Rightarrow m = 2 \] Now, compare trials 2 and 3 for \( H_2 \): \[ \left( \frac{0.244}{0.122} \right)^n = \frac{0.0678}{0.0339} \] Solve for \( n \): \[ 2^n = 2 \Rightarrow n = 1 \] Thus, the reaction is second-order with respect to NO and first-order with respect to \( H_2 \).

Step 2: Write the Rate Equation

Using the orders determined, write the rate equation as:\[ \text{Rate} = k[\text{NO}]^2[\text{H}_2]^1 \]

Step 3: Calculate the Rate Constant k

Using data from any trial, substitute into the rate equation. Use trial 1:\[ 0.136 = k(0.420)^2(0.122) \]Calculate \( k \): \[ k = \frac{0.136}{(0.420)^2 \times 0.122} \approx 6.18 \text{ L}^2/\text{mol}^2 \cdot \text{s} \]

Step 4: Find the Rate of Appearance of N2

Use the rate equation with the calculated \( k \) and the given concentrations:\[ \text{Rate} = 6.18 \times (0.350)^2 \times (0.205) \]Calculate:\[ \text{Rate} \approx 0.154 \text{ mol/L} \cdot \text{s} \]

Key Concepts

Reaction Order

The concept of reaction order is crucial in understanding how the concentration of reactants influences the rate of a chemical reaction. Reaction order is an exponent value in a rate equation, indicating how the rate is affected by the concentration of a particular reactant. Each reactant involved can have different orders, and these can be calculated by observing the change in rate with respect to the changes in reactant concentrations.

In the provided example, we study the reaction \[2 \text{NO}(g) + 2 \text{H}_2(g) \rightarrow \text{N}_2(g) + 2 \text{H}_2 \text{O}(g)\] by using experimental data. By comparing trials where one reactant concentration changes while the other remains constant, we determine that the reaction is second-order with respect to \(\text{NO}\) and first-order with respect to \(\text{H}_2\). This means, the rate of reaction is quite sensitive to changes in \([\text{NO}]\), increasing fourfold when \(\text{NO}\) is doubled, while the effect of \([\text{H}_2]\) is linear.

Rate Equation

The rate equation shows how the rate of reaction depends on the concentration of reactants. It is a mathematical expression that combines the reaction orders and the concentrations of reactants. Based on our findings, the rate equation for the given reaction is:\[ \text{Rate} = k [\text{NO}]^2 [\text{H}_2]^1 \]Here's what the equation tells us:

• k - This represents the rate constant, which will be discussed later. It is a unique value for a specific reaction at a given temperature.
• [\text{NO}]^2 - The concentration of \(\text{NO}\) raised to the power of 2. This indicates the reaction is second-order with respect to \(\text{NO}\), heavily influencing the reaction rate as its concentration changes.
• [\text{H}_2]^1 - The concentration of \(\text{H}_2\) raised to the power of 1, showing a first-order dependence.

Understanding the rate equation is fundamental in predicting how changes in conditions could affect the rate of reaction.

Rate Constant

The rate constant, denoted as \( k \), is a fundamental factor in the rate equation. It represents the proportionality constant that links the concentration of reactants to the rate of reaction. Importantly, the rate constant's value is specific to each reaction and depends on temperature.

To compute the rate constant, one can use any data point from the experiments and the rate equation. In this case, using trial 1: \[ 0.136 = k (0.420)^2 (0.122) \] Solving gives:\[ k \approx 6.18 \text{ L}^2/\text{mol}^2 \cdot \text{s} \] This value tells us how quickly the reaction proceeds when concentrations are in the specified units. A larger \( k \) implies a faster reaction under the same conditions. Understanding \( k \) provides insights into how factors like temperature shifts might affect reaction speed.

Rate of Reaction

The rate of reaction is a measure of how quickly a reactant is converted into products in a given time frame. It is often expressed in units of concentration per time, such as \( \text{mol/L} \cdot \text{s} \). For the reaction discussed, the rate is synonymous with the formation rate of \(\text{N}_2\).

To find the rate at specific reactant concentrations, we employ the rate equation:\[ \text{Rate} = 6.18 \times (0.350)^2 \times (0.205) \] Calculating this expression yields:\[ \text{Rate} \approx 0.154 \text{ mol/L} \cdot \text{s} \] This calculation shows that at these concentrations, \(\text{N}_2\) appears at a rate of 0.154 \( \text{mol/L} \cdot \text{s} \). Such understanding helps in controlling processes like synthesis in industrial settings by providing a means to predict the outcome of concentration changes.