Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 25

Consider the reaction $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ If, in a certain experiment, over a specific time period, 0.0048 mole of \(\mathrm{PH}_{3}\) is consumed in a \(2.0-\mathrm{L}\) container each second of reaction, what are the rates of production of \(\mathrm{P}_{4}\) and \(\mathrm{H}_{2}\) in this experiment?

Short Answer

Expert verified
The rates of production of P₄ and H₂ in this experiment are \(0.0012\ mol\cdot s^{-1}\) and \(0.0072\ mol\cdot s^{-1}\), respectively.

Step by step solution

01

Identify the balanced reaction equation

We are given the balanced reaction equation: $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$
02

Determine the consumption rate of PH₃

We are given the consumption rate of PH₃ as 0.0048 moles per second.
03

Determine the relationship between rates

Using the stoichiometry of the balanced reaction equation, we can write down the relationships of the rates of consumption and production: $$ \frac{\text{rate of consumption of PH}_{3}}{4} = \frac{\text{rate of production of P}_{4}}{1} = \frac{\text{rate of production of H}_{2}}{6} $$
04

Find the rate of production of P₄

Dividing the rate of consumption of PH₃ by 4 gives us the rate of production of P₄: $$ \text{rate of production of P}_{4} = \frac{\text{rate of consumption of PH}_{3}}{4} = \frac{0.0048\ mol\cdot s^{-1}}{4} = 0.0012\ mol\cdot s^{-1} $$
05

Find the rate of production of H₂

Multiplying the rate of production of P₄ by 6 gives us the rate of production of H₂: $$ \text{rate of production of H}_{2} = 6 \times \text{rate of production of P}_{4} = 6 \times 0.0012\ mol\cdot s^{-1} = 0.0072\ mol\cdot s^{-1} $$
06

Final Answer

The rates of production of P₄ and H₂ are 0.0012 mol/s and 0.0072 mol/s, respectively.

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.

Start your free trial

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 central idea of the collision model is that molecules must collide in order to react. Give two reasons why not all collisions of reactant molecules result in product formation.

The decomposition of NH3 to N2 and H2 was studied on two surfaces: Without a catalyst, the activation energy is 335 \(\mathrm{kJ} / \mathrm{mol}\) . a. Which surface is the better heterogeneous catalyst for the decomposition of \(\mathrm{NH}_{3} ?\) Why? b. How many times faster is the reaction at 298 \(\mathrm{K}\) on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor \(A\) is the same for each reaction. c. The decomposition reaction on the two surfaces obeys a rate law of the form $$ |text {Rate} =k \frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{H}_{2}\right]} $$ How can you explain the inverse dependence of the rate on the \(\mathrm{H}_{2}\) concentration?

Consider the general reaction $$ \mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC} $$ and the following average rate data over some time period \(\Delta t :\) $$ \begin{aligned}-\frac{\Delta \mathrm{A}}{\Delta t} &=0.0080 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \\\\-& \frac{\Delta \mathrm{B}}{\Delta t}=0.0120 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \\ \frac{\Delta \mathrm{C}}{\Delta t} &=0.0160 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ Determine a set of possible coefficients to balance this general reaction.

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M,\) con- centration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{min}^{-1}\) . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to $2.50 \times 10^{-3} M ?$

One mechanism for the destruction of ozone in the upper atmosphere is $\mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$ $\mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g)$ a. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{\mathrm{a}}\) for the uncatalyzed reaction $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) $$ is 14.0 \(\mathrm{kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed is 11.9 \(\mathrm{kJ}\) . What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C} ?\) Assume that the frequency factor \(A\) is the same for each reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation
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