91Ó°ÊÓ

Consider the following reaction between mercury(II) chloride and oxalate ion: $$ 2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) $$ The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the following rate data were obtained for the rate of disappearance of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}:\) \begin{tabular}{llll} \hline Experiment & {\(\left[\mathrm{HgCl}_{2}\right](M)\)} & {\(\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right](M)\)} & Rate \((M / \mathrm{s})\) \\ \hline 1 & 0.164 & 0.15 & \(3.2 \times 10^{-5}\) \\ 2 & 0.164 & 0.45 & \(2.9 \times 10^{-4}\) \\ 3 & 0.082 & 0.45 & \(1.4 \times 10^{-4}\) \\ 4 & 0.246 & 0.15 & \(4.8 \times 10^{-5}\) \\ \hline \end{tabular} (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) is \(0.25 \mathrm{M}\) if the temperature is the same as that used to obtain the data shown?

Step by step solution

01

Identify Parameters

To begin, identify the general form of the rate law: \[ \text{Rate} = k [\text{HgCl}_2]^x [\text{C}_2\text{O}_4^{2-}]^y \]where \( k \) is the rate constant, and \( x \) and \( y \) are the orders of the reaction with respect to \( \text{HgCl}_2 \) and \( \text{C}_2\text{O}_4^{2-} \) respectively.

02

Determine Reaction Orders

Use the experimental data to determine the orders of reaction. Compare Experiments 1 and 4 to find \( x \). The concentration of \( \text{C}_2\text{O}_4^{2-} \) remains constant, and doubling [\text{HgCl}_2] leads to rate \( 4.8 \times 10^{-5} \text{ M/s} / 3.2 \times 10^{-5} \text{ M/s} = 1.5 \). Compare Experiments 1 and 2 to find \( y \). Given [\text{HgCl}_2] is constant, tripling [\text{C}_2\text{O}_4^{2-}] leads to rate \( 2.9 \times 10^{-4} \text{ M/s} / 3.2 \times 10^{-5} \text{ M/s} = 9 \). Thus, \( x = 1 \) and \( y = 2 \).

03

Write Rate Law Equation

Substitute the reaction orders into the rate law: \[ \text{Rate} = k [\text{HgCl}_2]^1 [\text{C}_2\text{O}_4^{2-}]^2 \]Thus, the rate law is dependent on \( \text{HgCl}_2 \) to the first power and \( \text{C}_2\text{O}_4^{2-} \) to the second power.

04

Calculate Rate Constant

Using data from Experiment 1, substitute into the rate law to solve for \( k \):\[ 3.2 \times 10^{-5} = k (0.164)^1 (0.15)^2 \]\[ k = \frac{3.2 \times 10^{-5}}{0.164 \cdot 0.0225} \approx 0.00856 \, \text{M}^{-2}\text{s}^{-1} \]

05

Calculate Reaction Rate with New Concentrations

Using the rate constant found, substitute \( [\text{HgCl}_2] = 0.100 \, \text{M} \) and \([\text{C}_2\text{O}_4^{2-}] = 0.25 \, \text{M} \) into the rate law:\[ \text{Rate} = 0.00856 (0.100)^1 (0.25)^2 \]\[ \text{Rate} = 0.00856 \cdot 0.100 \cdot 0.0625 = 5.35 \times 10^{-5} \, \text{M/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Law

The rate law is an expression that quantifies the speed of a chemical reaction as a function of reactant concentrations. For the reaction between mercury(II) chloride and the oxalate ion, the rate law can be represented as \[ \text{Rate} = k [\text{HgCl}_2]^x [\text{C}_2\text{O}_4^{2-}]^y \]where: - \( k \) is the rate constant, - \( x \) and \( y \) are the reaction orders with respect to \( \text{HgCl}_2 \) and \( \text{C}_2\text{O}_4^{2-} \) respectively.The rate law indicates the dependence of the reaction rate on the concentrations of the reactants. It does not come directly from the stoichiometry of the overall reaction but is determined experimentally.In this example, through experimentation, it is found that \( x = 1 \) and \( y = 2 \), making the rate law:\[ \text{Rate} = k [\text{HgCl}_2]^1 [\text{C}_2\text{O}_4^{2-}]^2 \]This reveals that the reaction is first-order in \( \text{HgCl}_2 \) and second-order in \( \text{C}_2\text{O}_4^{2-} \). This means that doubling the concentration of \( \text{HgCl}_2 \) will double the rate, whereas doubling \( \text{C}_2\text{O}_4^{2-} \) increases the rate fourfold (\( 2^2 \)).

Reaction Order

The reaction order describes the power to which the concentration of a reactant is raised in the rate law. It provides insight into how changing a concentration affects the rate of reaction. In our example:- The reaction order for \( \text{HgCl}_2 \) is 1, indicating a linear relationship between this reactant’s concentration and the rate of reaction.- The reaction order for \( \text{C}_2\text{O}_4^{2-} \) is 2, meaning the rate is proportional to the square of its concentration.Steps to determine reaction orders involve comparing experiments where one reactant's concentration changes while the other's remains constant.For instance, keeping \( \text{C}_2\text{O}_4^{2-} \) constant, doubling \( \text{HgCl}_2 \) from one experiment to another reveals the order by how much the rate increases: it increased by a factor of 1.5 suggesting a first-order dependence. Similarly, tripling \( \text{C}_2\text{O}_4^{2-} \) while keeping \( \text{HgCl}_2 \) constant resulted in a ninefold increase in rate, indicating a second-order dependence.

Rate Constant

The rate constant \( k \) is a proportional factor in the rate law equation that provides information about the reaction speed. It varies with temperature and is determined experimentally.For this reaction, the rate constant can be found using experimental data. With the rate law known, substituting known concentrations and rates into the equation helps calculate \( k \).From Experiment 1, using \[ 3.2 \times 10^{-5} = k (0.164)^1 (0.15)^2 \]Simplifying, we find: \[ k = \frac{3.2 \times 10^{-5}}{0.164 \cdot 0.0225} \approx 0.00856 \, \text{M}^{-2}\text{s}^{-1} \]The units of \( k \) depend on the order of the overall reaction. Here, the overall reaction order is 3 (1 from \( \text{HgCl}_2 \) and 2 from \( \text{C}_2\text{O}_4^{2-} \)), yielding units of \( \text{M}^{-2}\text{s}^{-1} \). This clarifies that as concentrations increase, the rate increases dependent on these factors, underlining the significance of \( k \).

Concentration Effect

The effect of concentration on a reaction's rate is crucial in predicting how alterations in reactant quantities influence the speed of the reaction. In chemical kinetics, the rate law dictates this relationship.In our specific reaction, increasing the concentration of \( \text{HgCl}_2 \) results in a direct, proportional increase in the reaction rate, showcasing a first-order dependency—this means that doubling the concentration will double the reaction rate.For \( \text{C}_2\text{O}_4^{2-} \), the rate is more sensitive to changes. A third-order reaction implies the rate increases with the square of the concentration change. Therefore, a 50% increase in \( \text{C}_2\text{O}_4^{2-} \) concentration leads to a 125% increase in the reaction rate.To calculate the rate at new concentrations, such as \( [\text{HgCl}_2] = 0.100 \, \text{M} \) and \([\text{C}_2\text{O}_4^{2-}] = 0.25 \, \text{M} \), use:\[ \text{Rate} = 0.00856 \times (0.100) \times (0.25)^2 \]\[ \text{Rate} = 5.35 \times 10^{-5} \, \text{M/s} \]These calculations underscore how pivotal reactant concentrations are in controlling and predicting chemical reaction rates. They help chemists to optimize conditions for desired rates efficiently.