The rate law for the reaction $$ \begin{array}{c}{\mathrm{Cl}_{2}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \mathrm{HCl}(g)+\mathrm{CCl}_{4}(g)} \\ {\text { Rate }=k\left[\mathrm{Cl}_{2}\right]^{1 / 2}\left[\mathrm{CHCl}_{3}\right]}\end{array} $$ 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 \(k\) in the given rate law are: \(k \ [\text{units}] = \frac{\text{L}^{1/2}}{\text{mol}^{1/2}\cdot\text{s}}\).

Step by step solution

01

The given rate law for the reaction is: $$ \text{Rate} = k[\text{Cl}_2]^{1/2}[\text{CHCl}_3] $$ Here, k is the rate constant, and the concentrations of Cl₂ and CHCl₃ are raised to the power of 1/2 and 1, respectively. #Step 2: Write the units for each term#

We know that the rate has the unit of concentration per unit time, which in this case is mol/L·s. The concentrations in the rate law have the unit of mol/L. Rewrite the rate law with units: $$ \frac{\text{mol}}{\text{L}\cdot \text{s}} = k \left(\frac{\text{mol}}{\text{L}}\right)^{1/2} \left(\frac{\text{mol}}{\text{L}}\right) $$ #Step 3: Solve for the units of k#
02

Now, we will solve for the units of k by isolating k in the equation and cancelling out the units: $$ k = \frac{\frac{\text{mol}}{\text{L}\cdot \text{s}}}{\left(\frac{\text{mol}}{\text{L}}\right)^{1/2}\left(\frac{\text{mol}}{\text{L}}\right)} $$ $$ k = \frac{\frac{1}{\text{s}}}{\left(\frac{\text{mol}}{\text{L}}\right)^{1/2}} $$ $$ k = \frac{\text{L}^{1/2}}{\text{mol}^{1/2}\cdot\text{s}} $$ #Step 4: Present the final units for k#

The units for the rate constant k in the given rate law are: $$ k \ [\text{units}] = \frac{\text{L}^{1/2}}{\text{mol}^{1/2}\cdot\text{s}} $$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The decomposition of hydrogen iodide on finely divided gold at $150^{\circ} \mathrm{C}\( is zero order with respect to \)\mathrm{HI}$. The rate defined below is constant at $1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}$. $$ \begin{aligned} & 2 \mathrm{HI}(g) \stackrel{\text { ?u }}{\longrightarrow} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \\ \text { Rate }=-\frac{\Delta[\mathrm{HI}]}{\Delta t} &=k=1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ a. If the initial HI concentration was \(0.250 \mathrm{~mol} / \mathrm{L},\) calculate the concentration of HI at 25 minutes after the start of the reaction. b. How long will it take for all of the \(0.250 \mathrm{M}\) HI to decompose?

The decomposition of \(\mathrm{NO}_{2}(g)\) occurs by the following bimolecular elementary reaction: $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The rate constant at 273 \(\mathrm{K}\) is $2.3 \times 10^{-12} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\( , and the activation energy is 111 \)\mathrm{kJ} / \mathrm{mol}$ . How long will it take for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of 2.5 \(\mathrm{atm}\) to 1.5 \(\mathrm{atm}\) at \(500 . \mathrm{K}\) ? Assume ideal gas behavior.

Which of the following statement(s) is(are) true? a. The half-life for a zero-order reaction increases as the reaction proceeds. b. A catalyst does not change the value of \(\Delta \mathrm{E}\) . c. The half-life for a reaction, aA \(\longrightarrow\) products, that is first order in A increases with increasing \([\mathrm{A}]_{0} .\) d. The half-life for a second-order reaction increases as the reaction proceeds.

The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in A and to have a rate constant of $5.0 \times 10^{-2} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\( at \)25^{\circ} \mathrm{C}$ . An experiment was run at \(25^{\circ} \mathrm{C}\) where $[\mathrm{A}]_{0}=1.0 \times 10^{-3} \mathrm{M} .$ a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after $5.0 \times 10^{-3} \mathrm{s}\( has elapsed assuming \)[\mathrm{B}]_{0}=0$

Consider a reaction of the type aA \(\longrightarrow\) products, in which the rate law is found to be rate \(=k[\mathrm{A}]^{3}\) (termolecular reactions are improbable but possible). If the first half-life of the reaction is found to be \(40 .\) s, what is the time for the second half-life? Hint: Using your calculus knowledge, derive the integrated rate law from the differential rate law for a termolecular reaction: $$ \text {Rate} =\frac{-d[\mathrm{A}]}{d t}=k[\mathrm{A}]^{3} $$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free