Question

The reaction $$ 0^{\circ} \mathrm{C}, $$ These relationships hold only if there is a very small amount of \(\mathrm{I}_{3}^{-}\) present. What is the rate law and the value of the rate constant? (Assume that rate \(=-\frac{\Delta\left[\mathrm{H}_{2} \mathrm{SeO}_{3}\right]}{\Delta t} )\)

Short answer

The rate law for the reaction between hydrogen selenite and iodate is given by \(Rate = 2 \times 10^{-1}[\mathrm{IO_3^-}]\), and the value of the rate constant is \(2 \times 10^{-1} \frac{L}{mol\cdot min}\).

Step-by-step solution

Determine the balanced chemical equation

The balanced chemical equation for the reaction between hydrogen selenite and iodate can be written as follows: \[ 2\mathrm{IO_3}^- + 5\mathrm{H_2SeO_3} + 12\mathrm{H^+} \rightarrow I_3^- + 5\mathrm{Se} + 6\mathrm{H_2O} \] Step 2: Determine the order of the reaction with respect to each reactant

Determine the order of the reaction with respect to each reactant

To determine the order of the reaction with respect to \(\mathrm{IO_3^-}\), we will analyze the experimental data provided. We can see that when the concentration of \(\mathrm{IO_3^-}\) is doubled, the reaction rate doubles as well. This indicates that the reaction is first-order with respect to \(\mathrm{IO_3^-}\). The rate expression for \(\mathrm{IO_3^-}\) is given by: \[ Rate = k[\mathrm{IO_3^-}]^n \] Now, let's analyze the experimental data provided for \(\mathrm{H_2SeO_3}\). We can see that when the concentration of \(\mathrm{H_2SeO_3}\) is doubled, the reaction rate remains the same. This indicates that the reaction is zero-order with respect to \(\mathrm{H_2SeO_3}\). Thus, the rate law can be expressed as: \[ Rate = k[\mathrm{IO_3^-}]^n[\mathrm{H_2SeO_3}]^0 \] Step 3: Calculate the rate constant

Calculate the rate constant

Given that rate \(= -\frac{\Delta[\mathrm{H_2SeO_3}]}{\Delta t}\). We can use the experimental data and the rate expression we found in step 2 to calculate the rate constant. Let's use the data from the first row of the table: \(Rate = 4.0 \times 10^{-4} \frac{mol}{L \cdot min}\) \[\left[\mathrm{IO_3^-}\right] = 2.0 \times 10^{-3} \frac{mol}{L}\] Using the rate expression: \[Rate = k[\mathrm{IO_3^-}]^n\] We can plug in the values and find the rate constant: \[4.0\times10^{-4} = k(2.0\times10^{-3})^n\] Solving for k, we get: \[k = \frac{4.0\times 10^{-4}}{(2.0\times 10^{-3})}\] Thus, the rate constant, k, equals \(2 \times 10^{-1} \frac{L}{mol\cdot min}\). Step 4: Write the final rate law expression

Write the final rate law expression

Combining our findings from Steps 2 and 3, we can now write the complete rate law for the reaction: \[Rate = (2 \times 10^{-1} \frac{L}{mol\cdot min})[\mathrm{IO_3^-}]^1[\mathrm{H_2SeO_3}]^0\] Or simply as: \[Rate = 2 \times 10^{-1}[\mathrm{IO_3^-}]\] So, the rate law is \(Rate = 2 \times 10^{-1}[\mathrm{IO_3^-}]\) and the value of the rate constant is \(2 \times 10^{-1} \frac{L}{mol\cdot min}\).

Key concepts

Chemical Kinetics

Chemical kinetics focuses on understanding the speed or rate at which a chemical reaction occurs. It tells us how swiftly reactants are transformed into products, offering insights into the reaction mechanisms. When we study chemical kinetics, we often determine a rate law, like in the exercise given. The rate law provides a mathematical description of how the reaction rate depends on the concentration of reactants.

• Reaction rates are generally measured in terms of concentration changes over time, like moles per liter per second or minute.
• Kinetics can help us predict how a reaction will proceed under different conditions.
• It also offers clues about which reactants have the greatest impact on the rate of reaction.

Studying kinetics is crucial in fields like pharmaceuticals, where precise reaction control can impact drug production and effectiveness. Understanding kinetics can also help in environmental chemistry, aiding in the design of more effective chemical processes to reduce pollutants.

Reaction Order

Reaction order explains how the rate of a chemical reaction is affected by the concentration of reactants. It is usually expressed in the rate law as an exponent associated with each reactant concentration. In the given exercise, understanding the reaction order is key to figuring out the rate law.Here's a breakdown of reaction order:

• First-order reactions: The rate changes linearly with the concentration of a single reactant. If the concentration doubles, the rate doubles.
• Zero-order reactions: The rate is independent of the concentration of a reactant. No matter how much the concentration changes, the rate stays the same.

In the step-by-step solution, the reaction was determined to be first-order with respect to \[\mathrm{IO_3^-}\], and zero-order with respect to \[\mathrm{H_2SeO_3}\]. This decision was made by observing changes in the reaction rate when altering the concentrations.

Rate Constant

The rate constant, symbolized by \(k\), is a crucial element in the rate law. It is a coefficient reflecting the speed of a reaction, and it varies with conditions such as temperature. A rate constant links the reaction rate to the concentration of reactants through the rate law expression.

• The value of \(k\) is derived by inserting known values of concentration and rate into the rate law formula.
• The units of \(k\) vary depending on the overall order of the reaction.
• For a first-order reaction, \(k\) has units of \( ext{time}^{-1}\), while for a second-order reaction, \(k\) has units of \(\text{L/mol}\cdot\text{time}\).

In the provided exercise, the rate constant was calculated using the equation \[4.0\times 10^{-4} = k(2.0\times10^{-3})\]resulting in \(k\) equaling \[2 \times 10^{-1} \text{ L/mol}\cdot\text{min}.\]This illustrates how the rate constant connects the reaction rate to the concentration of \[\mathrm{IO_3^-}\], helping us understand the kinetic behavior of the reaction.