Problem 30 - Chemistry The Central Science In Si Units

Question

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law, \((\mathbf{b})\) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\) what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathbf{c})\) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of NO is increased to \(0.10 \mathrm{M}\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M} ?\) (d) What is the reaction rate at \(1000 \mathrm{~K}\) if \([\mathrm{NO}]\) is decreased to \(0.010 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to \(0.030 \mathrm{M} ?\)

Step-by-Step Solution

Answer

(b) 1.1025 M/s, (c) 6.0 M/s, (d) 0.18 M/s

Step 1: Write the Rate Law

Since the reaction is first order in \(\mathrm{H}_{2}\) and second order in \(\mathrm{NO}\), the rate law is written based on these orders. The rate \( R \) can be expressed as: \[ R = k [\mathrm{NO}]^2 [\mathrm{H}_2] \] where \( k \) is the rate constant.

Step 2: Calculate Reaction Rate for Part (b)

Given \( k = 6.0 \times 10^4 \mathrm{M}^{-2}\mathrm{s}^{-1} \), \([\mathrm{NO}]=0.035\mathrm{M}\), and \([\mathrm{H}_2]=0.015\mathrm{M}\), substitute these values into the rate law: \[ R = (6.0 \times 10^4) \times (0.035)^2 \times (0.015) \]. Calculate: \[ R \approx 6.0 \times 10^4 \times 0.001225 \times 0.015 = 1.1025 \ \mathrm{M} \ \mathrm{s}^{-1} \].

Step 3: Calculate Reaction Rate for Part (c)

Increase \([\mathrm{NO}]\) to 0.10 M and \([\mathrm{H}_2]\) is 0.010 M. Using the rate law: \[ R = (6.0 \times 10^4) \times (0.10)^2 \times (0.010) \]. Calculate: \[ R \approx 6.0 \times 10^4 \times 0.01 \times 0.01 = 6.0 \ \mathrm{M} \ \mathrm{s}^{-1} \].

Step 4: Calculate Reaction Rate for Part (d)

Decrease \([\mathrm{NO}]\) to 0.010 M and increase \([\mathrm{H}_2]\) to 0.030 M. Using the rate law: \[ R = (6.0 \times 10^4) \times (0.010)^2 \times (0.030) \]. Calculate: \[ R \approx 6.0 \times 10^4 \times 0.0001 \times 0.03 = 0.18 \ \mathrm{M} \ \mathrm{s}^{-1} \].

Key Concepts

Rate Law

The rate law is a mathematical expression that represents the relationship between the rate of a chemical reaction and the concentration of the reactants. Unlike a balanced chemical equation, which shows the stoichiometry of the reaction, a rate law must be determined experimentally. It provides a crucial understanding of how different concentration levels of reactants influence the speed of the reaction. In the given exercise, the rate law is represented as:

• \[ R = k [\mathrm{NO}]^2 [\mathrm{H}_2] \]
• Here, \(R\) is the rate of the reaction.
• \(k\) is the rate constant.
• \([\mathrm{NO}]^2\) shows that the rate is second order with respect to NO.
• \([\mathrm{H}_2]\) indicates that the rate is first order with respect to \(\mathrm{H}_2\).

The rate law provides insight into the molecularity of the reaction—revealing which reactants are directly involved in the rate-determining step.

Reaction Order

The reaction order is a key concept in reaction kinetics. It tells us how the rate of a chemical reaction depends on the concentration of the reactants. The exponents in the rate law equation represent the reaction order with respect to each reactant. Adding these exponents gives the overall reaction order.

• In the given example, the reaction is second order in \(\mathrm{NO}\) and first order in \(\mathrm{H}_2\).
• Thus, the overall order of this reaction is \(2 + 1 = 3\).

Understanding the reaction order can provide important clues about the reaction mechanism. For instance, a second-order dependence on NO suggests two NO molecules are involved in the rate-determining step. Meanwhile, the first-order dependence on \(\mathrm{H}_2\) suggests that only one \(\mathrm{H}_2\) molecule is involved.

Rate Constant

The rate constant, \(k\), is a proportionality constant in the rate law equation. It’s crucial because it relates the concentration of reactants to the reaction rate. The value of \(k\) is influenced by factors like temperature and the presence of a catalyst.

• For this reaction, \(k = 6.0 \times 10^4 \mathrm{M}^{-2}\mathrm{s}^{-1}\) at \(1000\, \mathrm{K}\).
• Notice that the units of \(k\) provide insight into the overall order of the reaction. For a third-order reaction, \(k\) will have units of \(\mathrm{M}^{-2}\mathrm{s}^{-1}\).
• The value of \(k\) is a constant at a fixed temperature but can change with varying reaction conditions.

Understanding \(k\) helps in predicting how the reaction rate will respond to changes in conditions and can assist in designing experiments to explore the reaction kinetics further.

Reaction Conditions

Reaction conditions refer to the environmental factors that affect the rate of a chemical reaction. Key conditions include concentration, temperature, and pressure.

• Concentration: The rate of reaction is dependent on the concentrations of reactants, as illustrated in the rate law. Using different concentrations for \([\mathrm{NO}]\) and \([\mathrm{H}_2]\) yielded differing reaction rates in the original exercise.
• Temperature: Increasing temperature typically increases reaction rates by providing more energy to overcome activation barriers. In our example, the given rate constant \(k\) applies specifically to a temperature of \(1000 \mathrm{~K}\).
• Catalysts: While not discussed in this exercise, catalysts lower the activation energy required, thus potentially altering \(k\) and enhancing reaction speed.

Considering these factors is essential when evaluating kinetic data and adapting conditions to control reaction rates effectively.