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Chapter 13: Problem 87

A reaction \(\mathrm{A}+\mathrm{B} \rightarrow\) Product is run in a balloon. (Both A and B are gases.) The balloon has a volume of 1 L and is initially loaded with 1 mole of \(\mathrm{A}\) and \(1 \mathrm{~mole}\) of \(\mathrm{B}\). The reaction has the rate law Rate \(=k[\mathrm{~A}]\) The reaction is run again using the same amount of reactants, but this time in a balloon that has a volume of \(0.5 \mathrm{~L}\). How much faster will the reaction proceed in the smaller balloon? Explain your answer.

Short Answer

Expert verified
The reaction will proceed 2 times faster in the smaller balloon. This is because when the volume is reduced to half, the concentration of reactant A doubles, and since the reaction rate depends on the concentration of A (Rate = k[A]), the overall rate of the reaction will be 2 times faster in the smaller balloon.

Step by step solution

01

Calculate the initial concentration of A in both cases

In both cases, we have 1 mole of A. We need to find the initial concentration of A in the balloons. For the first case, the balloon volume is 1 L: \[ [\mathrm{A}]_{initial_1} = \frac{1 \hspace{1mm} \mathrm{mole}}{1 \hspace{1mm} \mathrm{L}} = 1 \hspace{1mm} M \] For the second case, the balloon volume is 0.5 L: \[ [\mathrm{A}]_{initial_2} = \frac{1 \hspace{1mm} \mathrm{mole}}{0.5 \hspace{1mm} \mathrm{L}} = 2 \hspace{1mm} M \]
02

Calculate the reaction rates in both cases

Now that we have the initial concentrations of A, we can compute the reaction rates in both cases using the rate law. For the first case: \[ \mathrm{Rate_1} = k[\mathrm{A}]_{initial_1} = k(1 \hspace{1mm} M) \] For the second case: \[ \mathrm{Rate_2} = k[\mathrm{A}]_{initial_2} = k(2 \hspace{1mm} M) \] The ratio between the reaction rates is: \[ \frac{\mathrm{Rate_2}}{\mathrm{Rate_1}} = \frac{k(2 \hspace{1mm} M)}{k(1 \hspace{1mm} M)} \]
03

Determine how much faster the reaction proceeds in the smaller balloon

The last step is to simplify the calculated ratio to find how much faster the reaction will proceed in the smaller balloon. \[ \frac{\mathrm{Rate_2}}{\mathrm{Rate_1}} = \frac{2}{1} = 2 \] This means that the reaction will proceed 2 times faster in the smaller balloon. The reason for this is that when the volume is reduced to half, the concentration of reactant A doubles. As the reaction rate depends on the concentration of the reactant A (Rate = k[A]), the overall rate of the reaction will be 2 times faster in the smaller balloon.

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Most popular questions from this chapter

\(\Delta\) (Any quantity) is always defined as (final value of quantity) - (initial value of quantity). Now consider the quantity \(\Delta E_{\mathrm{rxn}}\). (a) For the forward reaction \(\mathrm{R} \rightarrow \mathrm{P}\), is \(\Delta E_{\mathrm{rxn}}=E_{\text {Reactants }}-E_{\text {Products }} ?\) Explain your answer. (b) According to your answer to (a), what does it mean when \(\Delta E_{\mathrm{rxn}}\) for a reaction is negative?

The slowest step in a reaction mechanism is called the _______-_______ step.

Explain the difference between a reaction intermediate and a catalyst in terms of the order in which each appears in the various steps of a reaction mechanism.

Chemical companies invest a considerable amount of time and energy in search of better catalysts for their chemical processes. Explain how this investment might pay off.

Why might one reaction have a much larger \(E_{a}\) than another reaction?
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