Problem 28 - Chemistry An Atoms First Approach

Question

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?

Step-by-Step Solution

Answer

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

Step 1: Write down the rate law expression

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

Step 2: 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. \( \frac{\text{mol}}{\text{L*s}} = k \left( \frac{\text{mol}}{\text{L}}\right)^{1 / 2}\left(\frac{\text{mol}}{\text{L}}\right) \)

Step 3: 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. \( k = \frac{\text{mol}}{\text{L*s}} \times \left(\frac{\text{L}}{\text{mol}}\right)^{1 / 2} \times \frac{\text{L}}{\text{mol}} \)

Step 4: 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: \( \frac{1}{\text{s} * \text{mol}^{\frac{1}{2}} * \text{L}^{\frac{1}{2}}} \)

Key Concepts

Reaction Kinetics

Understanding how quickly different chemical reactions occur can be very important in various fields. This is where reaction kinetics comes into play. Reaction kinetics is the study of the rates of chemical reactions and the factors that affect these rates. It tells us how fast reactants are converted into products.

There are several factors that influence reaction rates:

• Concentration of reactants: Generally, higher concentrations lead to higher reaction rates.
• Temperature: Increasing temperature usually speeds up a reaction.
• Catalysts: These substances can increase reaction rates without being consumed.

In reaction kinetics, we often use a rate law, which is an equation that connects the concentrations of reactants with the rate of a reaction. The exponents in the rate law indicate the order of the reaction with respect to each reactant. Overall, reaction kinetics helps us predict how a change in conditions can influence the speed of a chemical reaction.

Rate Constant

The rate constant, often represented as 'k', is a crucial element in the rate law of a chemical reaction. It provides a proportional link between the rate of reaction and the concentrations of reactants. Without it, the concentration terms in the rate law would not fit the actual speed of the reaction.

In the rate law, the rate constant gives us information about how fast a reaction proceeds under given conditions. Its value can be influenced by several factors, including temperature and the presence of a catalyst. For the given exercise, the units of the rate constant were determined by isolating 'k' in the rate law equation:

\[ k = \frac{1}{\text{s} * \text{mol}^{\frac{1}{2}} * \text{L}^{\frac{1}{2}}} \] This demonstrates the relationship between the rate of reaction and the concentration terms of the reactants. Understanding the rate constant helps chemists design and control chemical reactions better.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. Each chemical reaction has a distinct mechanism, which is the step-by-step sequence of elementary reactions by which overall chemical change occurs. These mechanisms underpin the rate laws that describe the reaction.

Using the given exercise as an example, the chlorination of chloroform involves a reaction between \( \mathrm{Cl}_{2}(g) \) and \( \mathrm{CHCl}_{3}(g) \). During this transformation, different bonds break and new bonds form, resulting in products \( \mathrm{HCl}(g) \) and \( \mathrm{CCl}_{4}(g) \). The stoichiometry of this particular reaction tells us the proportion of reactants that combine to form to the products.

Chemical reactions occur at different rates, and comprehending these rates aids in predicting how fast and to what extent the conversion will proceed. Understanding chemical reactions at a deeper level can also elucidate the dynamic interaction of molecules during the process.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us relate the quantities of reactants and products in a chemical reaction. It stems from the balanced chemical equation that represents a chemical reaction. Each coefficient in this equation indicates the relative number of moles of a given species.

For the reaction given in the exercise, knowing that one mole of \( \mathrm{Cl}_{2} \) reacts with one mole of \( \mathrm{CHCl}_{3} \) to form one mole of \( \mathrm{HCl} \) and one mole of \( \mathrm{CCl}_{4} \) allows us to determine how much of each substance is consumed or produced in the process.

Stoichiometry is vital for calculating reactant amounts needed or predicting amounts of products formed. This relationship not only aids in performing practical chemical calculations but also provides insight into the economy of reactions. It guides chemists in formulating precise proportions and enhances efficiency in industrial chemical processes.