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    Medium Rate Law Practice Questions

    April 2, 20269 min read1 views
    Medium Rate Law Practice Questions

    1. Concept Explanation

    A rate law is a mathematical equation that describes the relationship between the speed of a chemical reaction and the molar concentrations of its reactants. In chemical kinetics, the rate law expresses how the rate of reaction depends on the concentration of each reactant raised to a specific power, known as the reaction order. The general form for a reaction involving reactants A and B is written as: Rate = k[A]x[B]y, where "k" is the rate constant and "x" and "y" are the partial orders of the reaction. These orders are determined experimentally and cannot be predicted simply by looking at the stoichiometric coefficients of a balanced chemical equation.

    Understanding the Rate Law is essential for predicting how changes in concentration will affect the time it takes for a process to complete. The rate constant, k, is specific to a particular reaction at a given temperature; if the temperature changes, the value of k also changes, as described by the Arrhenius Equation. The sum of the individual orders (x + y) gives the overall reaction order, which provides insight into the reaction mechanism. For more foundational practice, you might find Reaction Order Practice Questions helpful in solidifying these concepts.

    According to Wikipedia's entry on Rate Equations, the rate law provides a link between the macroscopic change in concentration and the microscopic collisions between molecules. For many reactions, especially those with multiple steps, the rate law is dictated by the slowest step in the mechanism, known as the rate-determining step.

    2. Solved Examples

    Example 1: Determining the Rate Law from Initial Rates
    The following data were collected for the reaction 2A + B → C:
    - Exp 1: [A] = 0.10 M, [B] = 0.10 M, Rate = 2.0 x 10-4 M/s
    - Exp 2: [A] = 0.20 M, [B] = 0.10 M, Rate = 8.0 x 10-4 M/s
    - Exp 3: [A] = 0.10 M, [B] = 0.20 M, Rate = 4.0 x 10-4 M/s
    Find the rate law and the value of k.

    1. Compare Exp 1 and Exp 2: [A] doubles while [B] is constant. The rate increases by 4x (8.0/2.0). Since 22 = 4, the reaction is second order with respect to A.
    2. Compare Exp 1 and Exp 3: [B] doubles while [A] is constant. The rate increases by 2x (4.0/2.0). Since 21 = 2, the reaction is first order with respect to B.
    3. Write the rate law: Rate = k[A]2[B]1.
    4. Solve for k using Exp 1: 2.0 x 10-4 = k(0.10)2(0.10). k = 0.20 M-2s-1.

    Example 2: Units of the Rate Constant
    A reaction is found to be third-order overall. What are the units for the rate constant k if the rate is measured in M/s?

    1. The general formula for units of k is M(1-n) · t-1, where n is the overall order.
    2. Substitute n = 3: M(1-3) · s-1.
    3. Result: M-2s-1 (or L2/mol2·s).

    Example 3: Calculating Reaction Rate
    For the reaction Rate = (0.50 M-1s-1)[NO2]2, calculate the rate when the concentration of NO2 is 0.040 M.

    1. Identify the knowns: k = 0.50 M-1s-1 and [NO2] = 0.040 M.
    2. Plug values into the rate law: Rate = 0.50 × (0.040)2.
    3. Calculate: Rate = 0.50 × 0.0016 = 8.0 x 10-4 M/s.

    3. Practice Questions

    1. For the reaction A + B → products, doubling [A] while keeping [B] constant quadruples the rate. Doubling [B] while keeping [A] constant has no effect on the rate. Write the rate law.

    2. A reaction has the rate law: Rate = k[X][Y]2. If the concentration of X is tripled and the concentration of Y is halved, by what factor does the rate change?

    3. The rate constant for a first-order reaction is 3.5 x 10-3 s-1. If the initial concentration of the reactant is 0.25 M, what is the initial rate of the reaction?

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    4. Determine the overall reaction order for a reaction with a rate constant k = 1.2 x 10-2 M-1/2s-1.

    5. In a second-order reaction (Rate = k[A]2), the rate is 1.5 x 10-2 M/s when [A] = 0.10 M. What is the rate when [A] = 0.40 M?

    6. For the reaction 2NO + O2 → 2NO2, the rate law is Rate = k[NO]2[O2]. If the volume of the reaction container is reduced to half its original size, how will the rate change? (Hint: Concentration = n/V).

    7. A specific reaction is zero-order with respect to reactant A and second-order with respect to reactant B. If the concentration of A is doubled and B is tripled, what is the relative change in rate?

    8. Given the data for A + B → C:
    - [A]=0.1, [B]=0.1, Rate=4.0
    - [A]=0.2, [B]=0.1, Rate=16.0
    - [A]=0.1, [B]=0.2, Rate=8.0
    Calculate the value of the rate constant k with units.

    9. A reaction is found to follow the rate law Rate = k[A]x. When [A] is increased from 0.50 M to 1.50 M, the rate increases by a factor of 9. What is the value of x?

    10. Explain why the stoichiometric coefficients in the balanced equation 2N2O5 → 4NO2 + O2 do not necessarily match the exponents in its rate law.

    4. Answers & Explanations

    1. Rate = k[A]2
    Explanation: Doubling [A] results in a 4x rate increase (22=4), so order is 2 for A. Doubling [B] results in no change (20=1), so order is 0 for B. The rate law is Rate = k[A]2[B]0, which simplifies to Rate = k[A]2.

    2. Rate changes by a factor of 0.75 (or 3/4)
    Explanation: New Rate = k(3[X])(0.5[Y])2 = k(3[X])(0.25[Y]2) = 0.75 · k[X][Y]2. The rate is 0.75 times the original rate.

    3. 8.75 x 10-4 M/s
    Explanation: For a first-order reaction, Rate = k[Reactant]. Rate = (3.5 x 10-3 s-1)(0.25 M) = 8.75 x 10-4 M/s.

    4. 1.5 (or 3/2)
    Explanation: The units of k are M(1-n)s-1. Here, 1 - n = -1/2. Solving for n: n = 1 + 1/2 = 1.5.

    5. 0.24 M/s
    Explanation: In a second-order reaction, the rate is proportional to the square of the concentration. If [A] increases by 4x (0.10 to 0.40), the rate increases by 42 = 16. New Rate = 16 × (1.5 x 10-2) = 0.24 M/s.

    6. The rate increases by 8 times
    Explanation: Halving the volume doubles the concentration of all gaseous reactants. New Rate = k(2[NO])2(2[O2]) = k(4[NO]2)(2[O2]) = 8 · k[NO]2[O2].

    7. The rate increases by 9 times
    Explanation: Rate = k[A]0[B]2. Doubling A has no effect (20=1). Tripling B increases rate by 32 = 9. Total change = 1 × 9 = 9.

    8. k = 4000 M-2s-1
    Explanation: Order of A is 2 (rate quadruples when A doubles). Order of B is 1 (rate doubles when B doubles). Rate = k[A]2[B]. Using first row: 4.0 = k(0.1)2(0.1) → 4.0 = k(0.001) → k = 4000. Units for 3rd order are M-2s-1.

    9. x = 2
    Explanation: The concentration increased by a factor of 3 (1.50/0.50). The rate increased by a factor of 9. Since 3x = 9, x must be 2.

    10. Multi-step mechanisms
    Explanation: Stoichiometric coefficients represent the overall net change, but the rate law is determined by the elementary steps of the mechanism. Only for an elementary reaction do the coefficients match the orders. Complex reactions like this one involve intermediate steps that dictate the kinetic behavior.

    5. Quick Quiz

    Interactive Quiz 5 questions

    1. If a reaction is zero-order with respect to a reactant, what happens to the rate if the concentration of that reactant is tripled?

    • A The rate triples
    • B The rate increases ninefold
    • C The rate remains the same
    • D The rate decreases by one-third
    Check answer

    Answer: C. The rate remains the same

    2. What are the units of the rate constant for a second-order reaction?

    • A s-1
    • B M/s
    • C M-1s-1
    • D M-2s-1
    Check answer

    Answer: C. M-1s-1

    3. A reaction rate law is Rate = k[A][B]. What is the overall reaction order?

    • A Zero
    • B First
    • C Second
    • D Third
    Check answer

    Answer: C. Second

    4. Which factor generally increases the value of the rate constant k?

    • A Increasing the reactant concentration
    • B Increasing the temperature
    • C Decreasing the surface area
    • D Increasing the volume of the container
    Check answer

    Answer: B. Increasing the temperature

    5. In the rate law Rate = k[A]2, if the concentration of A is cut in half, the rate will:

    • A Decrease by a factor of 2
    • B Decrease by a factor of 4
    • C Remain unchanged
    • D Increase by a factor of 4
    Check answer

    Answer: B. Decrease by a factor of 4

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    6. Frequently Asked Questions

    How do you find the rate law from a balanced equation?

    You cannot determine the rate law solely from a balanced equation unless you are explicitly told the reaction is an elementary step. Rate laws must be determined experimentally by measuring how the initial rate changes with varying reactant concentrations.

    What is the difference between reaction rate and the rate constant?

    The reaction rate is the speed at which a reaction occurs (M/s) and changes as reactants are consumed. The rate constant (k) is a proportionality constant that is constant for a specific reaction at a specific temperature, regardless of concentration.

    Can reaction orders be fractions or negative numbers?

    Yes, while most introductory problems use whole numbers (0, 1, 2), reaction orders can be fractions or even negative if the mechanism involves complex equilibria or inhibition. These are typically discovered through rigorous chemical kinetics research.

    Why does temperature affect the rate constant k?

    Temperature increases the kinetic energy of molecules, leading to more frequent and more energetic collisions. This increases the fraction of collisions that overcome the activation energy, which is mathematically represented by an increase in the rate constant k.

    What does a zero-order reaction imply about the mechanism?

    A zero-order reaction implies that the rate is independent of the reactant's concentration, often occurring when the reaction is limited by a catalyst's surface area or light intensity rather than the amount of reactant available.

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