Question

The following set of data was obtained by the method of initial rates for the reaction: $$ \begin{array}{r} 2 \mathrm{HgCl}_{2}(\mathrm{aq})+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(\mathrm{aq}) \rightarrow \\ 2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{Hg}_{2} \mathrm{Cl}_{2} \end{array} $$ $$ \begin{array}{lll} \hline\left[\mathrm{HgCl}_{2}\right], \mathrm{M} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right], \mathrm{M}} & \text { Rate, } \mathrm{M} / \mathrm{s} \\ \hline 0.10 & 0.10 & 1.3 \times 10^{-7} \\ 0.10 & 0.20 & 5.2 \times 10^{-7} \\ 0.20 & 0.20 & 1.0 \times 10^{-6} \\ \hline \end{array} $$ What is the value of the rate constant, \(\mathrm{k}\) ? a. \(1.6 \times 10^{-4} 1 / \mathrm{M}^{2} \mathrm{~s}\) b. \(1.3 \times 10^{-4} 1 / \mathrm{M}^{2} . \mathrm{s}\) c. \(1.4 \times 10^{-7} 1 / \mathrm{M}^{2} . \mathrm{s}\) d. \(1.3 \times 10^{-6} 1 / \mathrm{M}^{2} \cdot \mathrm{s}\)

Short answer

The rate constant \( k \) is \( 1.3 \times 10^{-4} \text{ M}^{-2} \cdot \text{s}^{-1} \), corresponding to option b.

Step-by-step solution

Identify the Rate Law

The reaction is given as \(2 \text{HgCl}_2(\text{aq}) + \text{C}_2\text{O}_4^{2-}(\text{aq}) \rightarrow \). From the data, notice how changes in concentrations relate to changes in the rate. The general form of the rate law is \( \text{Rate} = k[\text{HgCl}_2]^m[\text{C}_2\text{O}_4^{2-}]^n \).

Determine the Reaction Orders

Using the first two experiments where [\( \text{HgCl}_2 \)] is constant, we calculate the change when [\( \text{C}_2\text{O}_4^{2-} \)] doubles: \[ n = \frac{\log(\frac{5.2 \times 10^{-7}}{1.3 \times 10^{-7}})}{\log(\frac{0.20}{0.10})} = 2 \]. Next, compare the second and third experiments where [\( \text{C}_2\text{O}_4^{2-} \)] is constant, and [\( \text{HgCl}_2 \)] doubles: \[ m = \frac{\log(\frac{1.0 \times 10^{-6}}{5.2 \times 10^{-7}})}{\log(\frac{0.20}{0.10})} = 1 \]. Thus, the rate law is \( \text{Rate} = k[\text{HgCl}_2]^1[\text{C}_2\text{O}_4^{2-}]^2 \).

Solve for the Rate Constant \( k \)

Select the first experiment (or any experiment) to substitute into the rate law: \( 1.3 \times 10^{-7} \text{ M/s} = k (0.10 \text{ M})^1 (0.10 \text{ M})^2 \). Simplifying the right side results in \( 0.0010 \text{ M}^3 \). Solving for \( k \): \[ k = \frac{1.3 \times 10^{-7}}{0.0010} = 1.3 \times 10^{-4} \text{ M}^{-2}\text{s}^{-1} \].

Match the Calculated \( k \) value with the Options

The calculated \( k \) is \( 1.3 \times 10^{-4} \text{ M}^{-2}\text{s}^{-1} \). Comparing with the given options, it matches with option b: \( 1.3 \times 10^{-4} \text{ M}^{-2} \cdot \text{s}^{-1} \).

Key concepts

Rate Law Determination

Understanding the rate law is crucial in reaction kinetics as it describes how the rate of a chemical reaction depends on the concentration of its reactants. The rate law expresses the relationship between the rate of a chemical reaction and the concentrations of the reactants. For a given reaction, the rate law can be written in the form:

• \( \text{Rate} = k[\text{A}]^m[\text{B}]^n \)

In this equation,

• \( k \) is the rate constant
• \([ \text{A} ] \) and \([ \text{B} ] \) are the concentrations of the reactants A and B
• \( m \) and \( n \) are the reaction orders for each reactant, respectively

The challenge lies in figuring out these unknowns: the rate constant \( k \) and the exponents \( m \) and \( n \), which denote the order of the reaction for each reactant. To establish the rate law, one needs experimental data showing how the rate of reaction varies with the concentration of reactants. This enables us to find the values of \( m \) and \( n \) by analyzing how changes in concentration affect the rate.

Reaction Order Calculation

Calculating the reaction order means determining the power to which each reactant concentration is raised in the rate law. This is done using experimental data and observing how changes in concentration affect reaction rates. For the examples provided:

• First, keep the concentration of one reactant constant and observe changes in the rate as the other reactant's concentration changes.
• Using the initial rate method, apply the formula for the reaction order: \[ n = \frac{\log(\text{Rate}_2/\text{Rate}_1)}{\log([\text{C}_2\text{O}_4^{2-}]_2/[\text{C}_2\text{O}_4^{2-}]_1)} \]

In our case, when \([\text{HgCl}_2]\) was held constant and \([\text{C}_2\text{O}_4^{2-}]\) was doubled, the rate increased fourfold. This indicates that the reaction is second order with respect to \(\text{C}_2\text{O}_4^{2-}\) as shown by \[ n = 2 \].
Next, when \([\text{C}_2\text{O}_4^{2-}]\) was constant and \([\text{HgCl}_2]\) was doubled, the rate doubled, suggesting a first-order reaction in terms of \(\text{HgCl}_2\) shown by \[ m = 1 \]. Calculating these orders helps us find the exponents for the rate law equation.

Initial Rate Method

The initial rate method is a valuable approach in determining the rate law of a reaction. This method relies on measuring the initial rate of reaction just after the reactants are mixed, preventing significant changes in their concentrations.

• Choose various experimental conditions to change concentrations of reactants while observing the initial rates.
• Analyze how these changes affect the speed of the reaction.

For example, consider the initial rate data from experiments where hydrogen chloride and oxalate concentrations are altered. The initial rate method involves:

• Calculating how the initial rate varies with concentration.
• Determining partial reaction orders by observing the relationship between initial concentrations and initial rates.

This process allows us to derive a quantitative measure of how sensitive the reaction rate is to changes in the concentration of each reactant. By examining ratio changes between concentrations and rates, we can confirm the order derived.

Rate Constant Calculation

The rate constant \( k \) is a critical factor in the rate law equation, reflecting the speed of a reaction without the influence of concentration changes. Calculating \( k \) involves substituting values from the rate law equation and known data from experiments.

• Plug the initial rate and concentrations from a chosen experiment into the rate law equation.
• Rearrange to solve for \( k \).

Using our example, from the rate law determined as \( \text{Rate} = k[\text{HgCl}_2]^1[\text{C}_2\text{O}_4^{2-}]^2 \), choose the initial data:

• Initial rate = \( 1.3 \times 10^{-7} \text{ M/s} \)
• Concentrations: \([\text{HgCl}_2] = 0.10 \text{ M} \), \([\text{C}_2\text{O}_4^{2-}] = 0.10 \text{ M} \)

Solve for \( k \): \[ k = \frac{1.3 \times 10^{-7}}{0.10^1 \times 0.10^2} = 1.3 \times 10^{-4} \text{ M}^{-2}\text{s}^{-1} \]This calculation is essential in understanding the reaction's characteristics and how fast it proceeds based on the given conditions.