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Chapter 14: Problem 39

For the reaction \(A \longrightarrow\) products, the data tabulated below are obtained. (a) Determine the initial rate of reaction (that is, \(-\Delta[\mathrm{A}] / \Delta t)\) in each of the two experiments. (b) Determine the order of the reaction. $$\begin{array}{ll} \hline \text { First Experiment } & \\ \hline[\mathrm{A}]=1.512 \mathrm{M} & t=0 \mathrm{min} \\ \begin{array}{l} | \mathrm{A}\rfloor=1.490 \mathrm{M} \\ {[\mathrm{A}]=1.469 \mathrm{M}} \end{array} & \begin{array}{l} t=1.0 \mathrm{min} \\ t=2.0 \mathrm{min} \end{array} \\ \hline & \\ \hline \text { Second Experiment } & \\ \hline[\mathrm{A}]=3.024 \mathrm{M} & t=0 \mathrm{min} \\ {[\mathrm{A}]=2.935 \mathrm{M}} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=2.852 \mathrm{M}} & t=2.0 \mathrm{min} \\ \hline \end{array}$$

Short Answer

Expert verified
The initial rates for the reactions are \(0.022\, M/min\) and \(0.086\, M/min\) respectively. The reaction is of second order.

Step by step solution

01

Determine the rate of reaction for the first experiment

The rate of reaction is calculated by the formula \(\Delta[\mathrm{A}]/ \Delta t\). Using the provided data for the first experiment, the rate of reaction is calculated: \[ -((1.469\, M - 1.512\, M)/(2.0\, min - 0\, min)) = 0.022\, M/min\]
02

Determine the rate of reaction for the second experiment

Using the same formula, and the provided data for the second experiment, calculate the rate of reaction: \[ -((2.852\, M - 3.024\, M)/(2.0\, min - 0\, min)) = 0.086\, M/min\].
03

Determine the order of the reaction

The order of a reaction is determined by observing how the rate changes when the concentration of the reactant is changed. From the rates, it can be seen that the reaction rate of Experiment 2 is almost 4 times that of Experiment 1 while the concentration of reactant A is twice as much. Therefore, it can be concluded that the reaction is of second order because when the concentration doubles, the rate quadruples.

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Most popular questions from this chapter

In three different experiments, the following results were obtained for the reaction \(A \longrightarrow\) products: \([\mathrm{A}]_{0}=1.00 \mathrm{M}, t_{1 / 2}=50 \mathrm{min} ;[\mathrm{A}]_{0}=200 \mathrm{M}, t_{1 / 2}=\) \(25 \min ;[\mathrm{A}]_{0}=0.50 \mathrm{M}, t_{1 / 2}=100 \mathrm{min} .\) Write the rate equation for this reaction, and indicate the value of \(k.\)

In the reaction \(A \longrightarrow\) products, 4.40 min after the reac- tion is started, \([\mathrm{A}]=0.588 \mathrm{M}\). The rate of reaction at this point is rate \(=-\Delta[\mathrm{A}] / \Delta t=2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1}.\) Assume that this rate remains constant for a short period of time. (a) What is \([\mathrm{A}] 5.00\) min after the reaction is started? (b) At what time after the reaction is started will \([\mathrm{A}]=0.565 \mathrm{M} ?\)

The following statements about catalysis are not stated as carefully as they might be. What slight modifications would you make in them? (a) A catalyst is a substance that speeds up a chemical reaction but does not take part in the reaction. (b) The function of a catalyst is to lower the activation energy for a chemical reaction.

In the reaction \(A \longrightarrow\) products, at \(t=0\), the \([\mathrm{A}]=0.1565 \mathrm{M} .\) After \(1.00 \mathrm{min},[\mathrm{A}]=0.1498 \mathrm{M},\) and after \(2.00 \mathrm{min},[\mathrm{A}]=0.1433 \mathrm{M}\) (a) Calculate the average rate of the reaction during the first minute and during the second minute. (b) Why are these two rates not equal?

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g),\) the total pressure increases while the partial pressure of \(\mathrm{A}(\mathrm{g})\) decreases. If the initial pressure of \(\mathrm{A}(\mathrm{g})\) in a vessel of constant volume is \(1.000 \times 10^{3} \mathrm{mmHg}\) (a) What will be the total pressure when the reaction has gone to completion? (b) What will be the total gas pressure when the partial pressure of \(\mathrm{A}(\mathrm{g})\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg} ?\)
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