Question

The rate law for the reaction $$ \mathrm{Cl}_{2}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \operatorname{HCl}(g)+\mathrm{CCl}_{4}(g) $$ 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 \(\mathrm{mol} / \mathrm{L}\) ?

Short answer

The units for the rate constant "k" in the given reaction are: \(\frac{\mathrm{L}^{\frac{3}{2}}}{\mathrm{mol}^{\frac{1}{2}} \cdot \mathrm{s}}\).

Step-by-step solution

Write down the given rate law equation and the units for concentration and time.

The rate law equation is given as: Rate = \(k[\mathrm{Cl}_{2}]^{\frac{1}{2}}[\mathrm{CHCl}_{3}]\) The units of concentration are mol/L, and the units of time are seconds.

Substitute the units of concentration and time into the rate law equation.

Replace the terms "Rate," "Cl2" and "CHCl3" with the concentration unit (mol/L) and "time" with the unit for time (s). The equation will now be: \( \frac{\mathrm{mol}}{\mathrm{L} \cdot\mathrm{s}} = k\left(\frac{\mathrm{mol}}{\mathrm{L}}\right)^{\frac{1}{2}} \left(\frac{\mathrm{mol}}{\mathrm{L}}\right)\)

Solve for the units of "k".

Now, we need to solve the equation to find the units of "k". First, multiply the exponents of the concentration units: \( \frac{\mathrm{mol}}{\mathrm{L} \cdot\mathrm{s}} = k\left(\frac{\mathrm{mol}}{\mathrm{L}}\right)^{\frac{1}{2}} \left(\frac{\mathrm{mol}}{\mathrm{L}}\right) = k\left(\frac{\mathrm{mol}^{\frac{1}{2}}}{\mathrm{L}^{\frac{1}{2}}}\right)\left(\frac{\mathrm{mol}}{\mathrm{L}}\right)\) Now, we simplify by multiplying the concentration units: \( \frac{\mathrm{mol}}{\mathrm{L} \cdot\mathrm{s}} = k\left(\frac{\mathrm{mol}^{\frac{3}{2}}}{\mathrm{L}^{\frac{3}{2}}}\right)\) Finally, solve for the units of "k": \( k = \frac{\mathrm{mol}}{\mathrm{L} \cdot\mathrm{s}} \times\frac{\mathrm{L}^{\frac{3}{2}}}{\mathrm{mol}^{\frac{3}{2}}} = \frac{\mathrm{L}^{\frac{3}{2}}}{\mathrm{mol}^{\frac{1}{2}} \cdot \mathrm{s}} \)

State the units of "k".

The units for the rate constant "k" in the given reaction are: \(\frac{\mathrm{L}^{\frac{3}{2}}}{\mathrm{mol}^{\frac{1}{2}} \cdot \mathrm{s}}\).

Key concepts

Rate Law

The rate law is an essential concept in chemical kinetics. It provides a mathematical relationship used to describe how the rate of a chemical reaction depends on the concentration of its reactants. For the reaction \( \mathrm{Cl}_{2}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \operatorname{HCl}(g)+\mathrm{CCl}_{4}(g), \) the rate law is given as \( \text{Rate} = k\left[\mathrm{Cl}_{2}\right]^{1/2}\left[\mathrm{CHCl}_{3}\right] \).
The rate equation indicates that the rate of the reaction is influenced by the concentrations of \(\mathrm{Cl}_2\) and \(\mathrm{CHCl}_3\). The presence of exponents within the rate law reveals that concentrations affect reaction rates differently.

• The concentration of \(\mathrm{Cl}_2\) is raised to the power of \(1/2\), showing a square root dependence.
• The concentration of \(\mathrm{CHCl}_3\) is linear or first order in its impact on the reaction rate.

This non-integral power of 1/2 for \(\mathrm{Cl}_2\) might suggest a more complex reaction mechanism than simple bimolecular collisions. Rate laws are vital because they help predict the speed at which a reaction proceeds under various conditions. They are determined experimentally and give insight into the steps involved in a reaction (mechanism).

Reaction Order

Reaction order reveals how the reaction rate is influenced by the concentration of each reactant. It is determined from the exponents of the concentrations in the rate law. In our reaction, the rate law is written as: \(\text{Rate} = k\left[\mathrm{Cl}_2\right]^{1/2}\left[\mathrm{CHCl}_3\right].\)

• The reaction is 1/2 order in \(\mathrm{Cl}_2\), as indicated by its exponent (1/2).
• It is first order in \(\mathrm{CHCl}_3\), since the exponent is 1.

The overall order of the reaction is the sum of these exponents, which is \(1/2 + 1 = 3/2\) or 1.5.
Understanding the reaction order is crucial in kinetics as it helps in predicting changes in the reaction rate for any given change in concentration. It also gives insights into the reaction's complexity, including possible intermediate steps not visible in the balanced chemical equation.

Rate Constant

The rate constant, symbolized by \(k\), is a proportionality factor in the rate law. It varies with the specific reaction and conditions such as temperature. For the given reaction \(\mathrm{Cl}_2(g)+\mathrm{CHCl}_3(g)\ \rightarrow \operatorname{HCl}(g)+\mathrm{CCl}_4(g)\), the rate law is \(\text{Rate} = k\left[\mathrm{Cl}_2\right]^{1/2}\left[\mathrm{CHCl}_3\right].\)
To find the units of \(k\), plug in the units of concentration (mol/L) and time (seconds) into the rate equation:
\[ \frac{\mathrm{mol}}{\mathrm{L} \cdot\mathrm{s}} = k\left(\frac{\mathrm{mol}}{\mathrm{L}}\right)^{1/2}\left(\frac{\mathrm{mol}}{\mathrm{L}}\right) \]
By simplifying, you get:
\[ k = \frac{\mathrm{L}^{3/2}}{\mathrm{mol}^{1/2} \cdot s} \]
The rate constant’s units change depending on the total order of the reaction. That's why figuring out the reaction order is important. Moreover, \(k\)'s value changes with temperature, supporting the rate's dependency on molecular energy and collision effectiveness. Understanding \(k\) allows chemists to adjust conditions optimally for desired reaction rates.