Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

The forward and reverse rate constants for the reac$\operatorname{tion} \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$ are $3.6 \times 10^{-3} / M \cdot \mathrm{s}$ and $8.7 \times 10^{-4} \mathrm{s}^{-1},$ respectively, at $323 \mathrm{K}$. Calculate the equilibrium pressures of all the species starting at $P_{\mathrm{A}}=1.6 \mathrm{atm}$ and $P_{\mathrm{B}}=0.44 \mathrm{atm}$

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

$$\begin{array}{l}P_{\mathrm{A}}=1.52 \mathrm{atm} \\P_{\mathrm{B}}=0.356 \mathrm{atm} \\P_{\mathrm{C}}=0.084 \mathrm{atm}\end{array}$$

Chemistry 102

Chapter 14

Chemical Equilibrium

University of Central Florida

Rice University

University of Kentucky

Lectures

10:03

In thermodynamics, a state of thermodynamic equilibrium is a state in which a system is in thermal equilibrium with its surroundings. A system in thermodynamic equilibrium is in thermal equilibrium, mechanical equilibrium, electrical equilibrium, and chemical equilibrium. A system is in equilibrium when it is in thermal equilibrium with its surroundings.

00:54

In chemistry, chemical equilibrium (also known as dynamic equilibrium) is a state of chemical stability in which the concentrations of the chemical substances do not change in the course of time due to their reaction with each other in a closed system. Chemical equilibrium is an example of dynamic equilibrium, a thermodynamic concept.

10:20

the forward and reverse ra…

04:53

The equilibrium constant $…

07:34

08:12

Consider the reaction:…

02:35

01:06

Given the reaction at a ce…

04:45

there are multiple steps involved in this problem. First, we need to figure out what the equilibrium constant is. Equilibrium constant is typically just a ratio of the rate constants in the forward direction, divided by the rate constant in the reverse direction. So if this is our generic reaction than K is going to be equal to the pressure of sea, or KP in particular is going to be equal to the pressure of sea over the pressure of a multiplied by the pressure of B de que value. Or, more specifically, the case see value is in the ratio of the rate constant in the forward direction, divided by the rate constant in the reverse direction. This is Casey because rate constants are a function of concentrations, so K C then is 4.14 now to determine the final pressures at equilibrium. We need a K p value because we are given initial pressures and we're working with pressures, not a Casey value. So we need to convert the K C value into the KP value. Before we do that, let's figure out what the equilibrium pressures would be in terms of an unknown there in terms of the variable X. If we have start with a 1.6 atmosphere of A and I will 0.44 atmospheres of B and nothing to see that it has to shift to the right in order to establish equilibrium because we can't have any pressures at zero. So the B minus X minus X and plus X because all of the coefficients are one, then it equilibrium will have 1.6 minus X for a 0.44 minus x for B and just X for C. So now let's go back to calculate in our KP value from RKC value, KP is going to be equal to Casey multiplied by our, which is 0.8 to 1 as a constant multiplied by t, which they gave to us in Kelvin at 3 23 All of that raised to the Delta End and Delta N is going to be one full of gas minus two moles of gas or negative one. So when we calculate our KP value, recognizing that this is now raised to the negative one, we get 0.156 Now that we know are KP value and we know our equilibrium concentrations while equilibrium pressures expressed in terms of the variable X we can then solve for X. Our expression will then be the KP value is going to be equal to the pressure of C, which is X divided by the pressures of A and B in the denominator multiplied by each other, which will be for a 1.6 minus X and for B 0.44 minus X, then to save some space. And some time I'll just show you the answer to X. Once you perform the algebra, the algebra gives us an X value of 0.84 Now that we know the X value than solving for the the pressures of A B and C at equilibrium is pretty straightforward, See is just going to be equal to X, which will be 0.84 a will be 1.6 minus X or 1.52 and then see, I'm sorry. B will be point for four minus X or 40.356 atmospheres

View More Answers From This Book

Find Another Textbook

Numerade Educator

04:04

A bond is usually more ionic than covalent in characterwhen?A. there…

01:40

The addition of potassium chromate to a solution of the slightlysoluble …

03:56

Which of the following would produce a reaction?HCl(aq) + Ag(s)H…

08:04

a. An ideal gas has a volume of 18.25 L at a temperature of 15.9°C. The …

04:19

Calculate the molarity of each of the following solutions.13.20 mg KI in…

04:38

A 14.5g sample of anhydrous sodium sulfate, Na(2)SO(4), isdissolved in 4…

08:51

In 1986 an electrical power plant in Taylorsville, Georgia,burned 8,376,…

01:56

Provide the proper coefficients (including 1's) required tobalance …

Which of the following accurately depicts the position ofelectrons in an…

it is found that at 26C and 1.16 atm pressure, 2.64 L of gas hasa mass o…