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Chapter 11: Problem 28

The rate law for the reaction $$\mathrm{Cl}_{2}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \mathrm{HCl}(g)+\mathrm{CCl}_{4}(g)$$ is $$\text { Rate }=k\left[\mathrm{Cl}_{2}\right]^{1 / 2}\left[\mathrm{CHCl}_{3}\right]$$ What are the units for \(k\), assuming time in seconds and concentration in mol/L?

Short Answer

Expert verified
The units for the rate constant $k$ are: \( \frac{1}{\text{s} * \text{mol}^{\frac{1}{2}} * \text{L}^{\frac{1}{2}}} \)

Step by step solution

01

Write down the rate law expression

The rate law expression for the given reaction is: \( \text { Rate }=k\left[\mathrm{Cl}_{2}\right]^{1 / 2}\left[\mathrm{CHCl}_{3}\right] \)
02

Substitute the units for each term in the expression

Now we need to substitute the units for the rate, concentration of Cl2, and concentration of CHCl3 in the rate law expression. \( \frac{\text{mol}}{\text{L*s}} = k \left( \frac{\text{mol}}{\text{L}}\right)^{1 / 2}\left(\frac{\text{mol}}{\text{L}}\right) \)
03

Rearrange the expression to solve for the units of k

In this step, we will isolate k in the expression, which will give us the units for k. \( k = \frac{\text{mol}}{\text{L*s}} \times \left(\frac{\text{L}}{\text{mol}}\right)^{1 / 2} \times \frac{\text{L}}{\text{mol}} \)
04

Simplify the expression for the units of k

In this step, we will simplify the expression to find the units of k. \( k = \frac{\text{mol}}{\text{s}} \times \left(\frac{1}{\text{mol}^{\frac{1}{2}}}\right) \times \frac{1}{\text{L}^{\frac{1}{2}}}\) Now, we can rewrite the expression as: \( k = \frac{1}{\text{s} * \text{mol}^{\frac{1}{2}} * \text{L}^{\frac{1}{2}}} \) Hence, the units for the rate constant k are: \( \frac{1}{\text{s} * \text{mol}^{\frac{1}{2}} * \text{L}^{\frac{1}{2}}} \)

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Most popular questions from this chapter

How does temperature affect \(k,\) the rate constant? Explain.

The reaction $$\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ was studied by performing two experiments. In the first experiment the rate of disappearance of NO was followed in the presence of a large excess of \(\mathrm{O}_{3}\). The results were as follows \(\left(\left[\mathrm{O}_{3}\right]\right.\) remains effectively constant at \(1.0 \times 10^{14}\) molecules/cm \(^{3}\) ): In the second experiment [NO] was held constant at \(2.0 \times 10^{14}\) molecules/cm \(^{3}\). The data for the disappearance of \(\mathbf{O}_{3}\) are as follows: a. What is the order with respect to each reactant? b. What is the overall rate law? c. What is the value of the rate constant from each set of experiments? $$\text { Rate }=k^{\prime}[\mathrm{NO}]^{x} \quad \text { Rate }=k^{\prime \prime}\left[\mathrm{O}_{3}\right]^{y}$$ d. What is the value of the rate constant for the overall rate law? $$\text { Rate }=k[\mathrm{NO}]^{\mathrm{x}}\left[\mathrm{O}_{3}\right]^y$$

Chemists commonly use a rule of thumb that an increase of \(10 \space\mathrm{K}\) 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 \(35^{\circ} \mathrm{C} ?\)

Table \(11-2\) 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?

Each of the statements given below is false. Explain why. a. The activation energy of a reaction depends on the overall energy change \((\Delta E)\) for the reaction. b. The rate law for a reaction can be deduced from examination of the overall balanced equation for the reaction. c. Most reactions occur by one-step mechanisms.
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