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The following data were obtained for the reaction \(2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) where \(\quad\) Rate \(=-\frac{\Delta\left[\mathrm{ClO}_{2}\right]}{\Delta t}\)

Short Answer

Expert verified
The given reaction is already balanced. To calculate the rate of the reaction, we need the initial and final concentrations of the reactant \(\mathrm{ClO}_{2}\) and the time interval \(\Delta t\). Once you have the necessary data, use the given rate expression: \[Rate = -\frac{\Delta\left[\mathrm{ClO}_{2}\right]}{\Delta t}\] Plug in the values and calculate the rate of the reaction. Since we do not have the concentrations and time interval, we cannot provide a numerical value for the rate.

Step by step solution

01

Rewrite the reaction in a balanced form

Write the given reaction in a balanced form: \[2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\]
02

Determine the relationship between the concentration of the reactants and the rate of the reaction

The rate of the reaction is given as: \[Rate =-\frac{\Delta\left[\mathrm{ClO}_{2}\right]}{\Delta t}\]
03

Use the given data to calculate the rate of the reaction

Based on the given exercise, no additional data is provided, so we cannot directly calculate the rate of reaction. It's crucial to have the initial and final concentrations of the reactant \(\mathrm{ClO}_{2}\) and the time interval over which the change occurred, expressed as\(\Delta t\). However, once you have these values, plug them into the given Rate expression: \[Rate = -\frac{\Delta\left[\mathrm{ClO}_{2}\right]}{\Delta t}\] And perform the necessary calculation to find the rate of the reaction.

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Key Concepts

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

Reaction Rate
The reaction rate is essentially a measure of how fast or slow a chemical reaction occurs. When it comes to understanding reaction rates, it's important to grasp that they can be determined by following changes in the concentrations of reactants or products over time. For our specific reaction, the rate is described in terms of the disappearance of chlorite ions, \[Rate = -\frac{\Delta[\mathrm{ClO}_2]}{\Delta t}\]This equation tells us that the rate of reaction is calculated by identifying the change in concentration of \( \mathrm{ClO}_2 \) over a particular time interval, divided by the length of that interval. If the concentration of \( \mathrm{ClO}_2 \) decreases quickly, the reaction proceeds at a higher rate; if it decreases slowly, the reaction rate is lower.
  • Negative sign: Indicates that the concentration of \( \mathrm{ClO}_2 \) is decreasing.
  • \( \Delta \): Represents a change in a quantity, here it means a change in concentration and time.
Balanced Chemical Equation
A balanced chemical equation provides a detailed depiction of the substances' relative amounts involved in a chemical reaction. For our exercise, the balanced equation is:\[2 \mathrm{ClO}_2(aq) + 2 \mathrm{OH}^-(aq) \longrightarrow \mathrm{ClO}_3^-(aq) + \mathrm{ClO}_2^-(aq) + \mathrm{H}_2O(l)\]Balancing a chemical equation is vital because it reflects the law of conservation of mass, indicating that the amount of each element must be the same on both sides of the equation.
  • Coefficients: Numbers before the formulas (like '2' in front of \( \mathrm{ClO}_2 \)) showing the exact amount of molecules needed.
  • Subscripts: Small numbers within the chemical formula, which should not be changed when balancing equations.
A balanced equation ensures that the same number of each type of atom appears on both the left side (reactants) and right side (products) of the equation.
Concentration Change
Concentration change is a vital aspect of determining reaction rates in chemical kinetics. In this context, concentration refers to the amount of substance present in a given volume of solution. For our chemical reaction, the concentration change concerns \( \mathrm{ClO}_2 \) as a key reactant. This change over time is expressed as \( \Delta[\mathrm{ClO}_2] \).During the reaction, the concentration of \( \mathrm{ClO}_2 \) decreases, which can be measured at different time points. By calculating the difference between the initial and final concentrations, one can figure out the concentration change. This concept highlights:
  • Initial and final concentrations: The concentration at the beginning and end of a measured time period.
  • Time interval (\( \Delta t \)): The duration over which the reaction is observed.
Understanding the concentration change is crucial for practical applications, such as chemical manufacturing, where precise control over reaction rates is required.
Rate Law
The rate law provides a mathematical relationship that expresses how the reaction rate depends on the concentration of the reactants. Although the exercise doesn't explicitly cover the rate law, understanding it is helpful for interpreting how reaction rates are affected by reactant concentrations.A general form of a rate law can be expressed as:\[Rate = k [A]^m [B]^n\]Here, \( k \) is the rate constant, and \( m \) and \( n \) are the reaction order with respect to each reactant (\( [A] \) and \( [B] \)).
  • Rate constant (\( k \)): A specific constant value for each reaction at a given temperature.
  • Reaction order: Exponents \( m \) and \( n \), typically determined experimentally, illustrating how the rate depends on the concentration of each reactant.
Understanding rate laws allows chemists to predict how changes in conditions, such as reactant concentrations, will impact the reaction speed, facilitating effective control of industrial and laboratory processes.

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Most popular questions from this chapter

Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\), with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{array}{l} k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{~s}^{-1} \\ k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{~s}^{-1} \end{array} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) \(4.00[\mathrm{~B}] ?\)

The rate law for the reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ at some temperature is $$ \text { Rate }=-\frac{\Delta[\mathrm{NOBr}]}{\Delta t}=k[\mathrm{NOBr}]^{2} $$ a. If the half-life for this reaction is \(2.00 \mathrm{~s}\) when \([\mathrm{NOBr}]_{0}=\) \(0.900 \mathrm{M}\), calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of NOBr to decrease to \(0.100 \mathrm{M}\) ?

For the reaction \(\mathrm{A} \rightarrow\) products, successive half-lives are observed to be \(10.0,20.0\), and \(40.0\) min for an experiment in which \([\mathrm{A}]_{0}=\) \(0.10 \mathrm{M} .\) Calculate the concentration of \(\mathrm{A}\) at the following times. a. \(80.0 \mathrm{~min}\) b. \(30.0 \mathrm{~min}\)

One mechanism for the destruction of ozone in the upper atmosphere is $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & \text { Slov } \\ \mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \text { Fast } \\ \hline \end{array} $$ Overall reaction \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g})\) a. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{\mathrm{a}}\) for the uncatalyzed reaction $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2} $$ is \(14.0 \mathrm{~kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed is \(11.9 \mathrm{~kJ}\). What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C}\) ? Assume that the frequency factor \(A\) is the same for each reaction.

Chemists commonly use a rule of thumb that an increase of \(10 \mathrm{~K}\) in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to \(35^{\circ} \mathrm{C}\) ?

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