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Chapter 12: Problem 32

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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Most popular questions from this chapter

Draw a rough sketch of the energy profile for each of the following cases: a. \(\Delta E=+10 \mathrm{~kJ} / \mathrm{mol}, E_{\mathrm{a}}=25 \mathrm{~kJ} / \mathrm{mol}\) b. \(\Delta E=-10 \mathrm{~kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{~kJ} / \mathrm{mol}\) c. \(\Delta E=-50 \mathrm{~kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{~kJ} / \mathrm{mol}\)

The central idea of the collision model is that molecules must collide in order to react. Give two reasons why not all collisions of reactant molecules result in product formation.

Table \(12.2\) illustrates how the average rate of a reaction decreases with time. Why does the average rate decrease with time? How does the instantaneous rate of a reaction depend on time? Why are initial rates used by convention?

Consider the reaction $$ 3 \mathrm{~A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E} $$ where the rate law is defined as $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{~A}]^{2}[\mathrm{~B}][\mathrm{C}] $$ An experiment is carried out where \([\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \mathrm{M}\) and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} M\) a. If after \(3.00 \mathrm{~min},[\mathrm{~A}]=3.26 \times 10^{-5} M\), calculate the value of \(k\). b. Calculate the half-life for this experiment. c. Calculate the concentration of \(\mathrm{B}\) and the concentration of \(\mathrm{A}\) after \(10.0 \mathrm{~min}\).

The decomposition of hydrogen iodide on finely divided gold at \(150^{\circ} \mathrm{C}\) is zero order with respect to HI. The rate defined below is constant at \(1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) $$ \begin{array}{r} 2 \mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \\ \text { Rate }=-\frac{\Delta[\mathrm{HI}]}{\Delta t}=k=1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{array} $$ a. If the initial HI concentration was \(0.250 \mathrm{~mol} / \mathrm{L}\), calculate the concentration of HI at 25 minutes after the start of the reaction. b. How long will it take for all of the \(0.250 M\) HI to decompose?
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