Hard Rate Law Practice Questions

Concept Explanation

A rate law is a mathematical equation that describes the relationship between the concentration of reactants and the speed of a chemical reaction. Unlike stoichiometry, the exponents in a rate law (the reaction orders) must be determined experimentally rather than from the balanced chemical equation. For a general reaction aA + bB → Products, the rate law is expressed as Rate = k[A]x[B]y, where k is the rate constant, and x and y are the partial orders of the reaction. Understanding reaction order practice questions is essential for mastering these complex calculations.

At an advanced level, rate laws involve multi-step mechanisms, the steady-state approximation, and the influence of temperature as described by the Arrhenius equation. In these systems, the "bottleneck" or slowest step determines the overall rate of the reaction. However, when the slow step is preceded by fast equilibrium steps, the rate law must be derived by substituting intermediate concentrations with reactant concentrations. This frequently requires a deep understanding of Arrhenius equation practice questions to account for how the rate constant k changes with environmental conditions.

Solved Examples

Below are three worked examples of Hard Rate Law Practice Questions to demonstrate the application of the steady-state approximation and experimental data analysis.

1. Example 1: Determining Order from Initial Rates
   Consider the reaction: 2NO(g) + Cl2(g) → 2NOCl(g). Experimental data shows that doubling [NO] quadruples the rate, while doubling [Cl2] doubles the rate. Find the rate law.

   1. Identify the effect of [NO]: Since 2x = 4, x = 2. The reaction is second order with respect to NO.

   2. Identify the effect of [Cl2]: Since 2y = 2, y = 1. The reaction is first order with respect to Cl2.

   3. Combine into the rate law: Rate = k[NO]2[Cl2].

2. Example 2: Mechanism with a Fast Initial Step
   Reaction: 2NO + O2 → 2NO2. Mechanism: (1) NO + NO ⇌ N2O2 (fast equilibrium); (2) N2O2 + O2 → 2NO2 (slow).

   1. Write the rate for the slow step: Rate = k2[N2O2][O2].

   2. Since N2O2 is an intermediate, use the equilibrium constant K = [N2O2] / [NO]2.

   3. Rearrange for the intermediate: [N2O2] = K[NO]2.

   4. Substitute back into the rate law: Rate = k2K[NO]2[O2], or Rate = k'[NO]2[O2].

3. Example 3: Calculating the Rate Constant k
   Given the rate law Rate = k[A][B]2, if the rate is 1.5 x 10-3 M/s when [A] = 0.1 M and [B] = 0.2 M, find k.

   1. Rearrange the formula: k = Rate / ([A][B]2).

   2. Substitute values: k = (1.5 x 10-3) / (0.1 * 0.22).

   3. Calculate: k = (1.5 x 10-3) / (0.004) = 0.375 M-2s-1.

Practice Questions

Test your knowledge with these Hard Rate Law Practice Questions. These are designed to challenge your understanding of kinetics and mechanism derivation.

1. A reaction is found to be 3/2 order with respect to reactant A and -1 order with respect to reactant B. If the concentration of A is doubled and the concentration of B is tripled, by what factor does the reaction rate change?

2. The decomposition of ozone (2O3 → 3O2) follows a mechanism where O3 ⇌ O2 + O (fast equilibrium) and O + O3 → 2O2 (slow). Derive the rate law for this reaction.

3. For the reaction A + B → C, the following data was collected: [A]=0.1, [B]=0.1, Rate=0.02; [A]=0.2, [B]=0.1, Rate=0.08; [A]=0.1, [B]=0.2, Rate=0.02. Determine the full rate law including the value of k.

4. The rate constant for a second-order reaction is 8.0 x 10-2 M-1s-1. If the initial concentration is 0.50 M, how long will it take for the concentration to decrease to 0.10 M? (Hint: Use integrated rate laws found in our rate law practice questions guide).

5. Explain why the rate of a reaction might decrease as the concentration of a product increases, citing a specific rate law structure.

6. A reaction A + B → C has the rate law Rate = k[A]2[B]. If the volume of the reaction vessel is halved, by what factor does the rate increase?

7. Using the steady-state approximation, derive the rate law for the formation of C in the mechanism: A ⇌ B (k1, k-1) and B + D → C (k2), where B is a highly reactive intermediate.

8. A reaction has an activation energy of 50 kJ/mol. By what factor does the rate constant increase if the temperature is raised from 300 K to 310 K?

9. Determine the units of the rate constant for a reaction with an overall order of 2.5.

10. In a complex reaction, the rate law is Rate = k[H2][I2] / [HI]. If the concentration of HI is quadrupled while others remain constant, what happens to the rate?

Answers & Explanations

1. Factor = (2)1.5 * (3)-1. 21.5 is approximately 2.828. Multiplying by 1/3 gives 0.943. The rate decreases slightly to about 94.3% of its original value.

2. Rate = k[O3]2[O2]-1. From the slow step: Rate = k2[O][O3]. From equilibrium: K = [O2][O] / [O3] → [O] = K[O3]/[O2]. Substitution yields the final law.

3. Rate = 2.0 M-1s-1 [A]2. Comparing trials 1 and 2: [A] doubles, rate quadruples (Order=2). Comparing 1 and 3: [B] doubles, rate is constant (Order=0). k = 0.02 / (0.1)2 = 2.0.

4. 100 seconds. For second order: 1/[A]t - 1/[A]0 = kt. 1/0.1 - 1/0.5 = (0.08)t → 10 - 2 = 0.08t → 8 = 0.08t → t = 100.

5. This occurs in inhibited reactions where a product is involved in a fast reverse step of the mechanism, appearing in the denominator of the rate law (e.g., Rate = k[A]/[P]).

6. 8 times. Halving volume doubles concentrations. Rate' = k(2[A])2(2[B]) = 8 * k[A]2[B].

7. Rate = k2k1[A][D] / (k-1 + k2[D]). Set d[B]/dt = 0 = k1[A] - k-1[B] - k2[B][D]. Solve for [B] and substitute into Rate = k2[B][D].

8. Approximately 1.9. Use ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). ln(k2/k1) = (50000/8.314)(1/300 - 1/310) ≈ 0.647. e0.647 ≈ 1.91.

9. M-1.5s-1. Formula for units is M1-ns-1. 1 - 2.5 = -1.5.

10. Rate decreases by a factor of 4. Since [HI] is in the denominator with an exponent of 1, the rate is inversely proportional to [HI].

Quick Quiz

6. Frequently Asked Questions

What makes a rate law question "hard"?

Hard questions typically involve non-integer reaction orders, multi-step mechanisms requiring the steady-state approximation, or integrated rate laws for reactions that are not first-order. They may also require you to link kinetic data with thermodynamic properties like activation energy.

Can a reaction order be negative?

Yes, a negative reaction order occurs when a species (often a product or an intermediate) inhibits the reaction. This means that increasing the concentration of that species actually slows down the overall rate of the reaction.

How do I identify the rate-determining step in a mechanism?

The rate-determining step is the slowest step in a proposed mechanism. It acts as a bottleneck, and the overall rate law of the reaction must match the stoichiometry of this step, potentially after substituting for any intermediates.

What is the difference between a differential and an integrated rate law?

A differential rate law expresses the reaction rate as a function of concentration at a specific moment. An integrated rate law expresses the concentration of reactants as a function of time, allowing you to calculate how much reactant remains after a certain duration.

Why is the rate constant k temperature-dependent?

The rate constant depends on temperature because temperature determines the kinetic energy of the molecules. According to the Transition State Theory, higher temperatures increase the frequency of collisions and the fraction of molecules with enough energy to overcome the activation barrier.

How do I handle reactions with multiple reactants?

To determine the order of multiple reactants, use the method of initial rates by changing the concentration of one reactant at a time while keeping others constant. If the rate law is already known, you can calculate the effect of simultaneous changes by multiplying the individual factors together.

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