Download the App!

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

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

Consider the following equilibrium process at $686^{\circ} \mathrm{C}$$$\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$The equilibrium concentrations of the reacting species are $[\mathrm{CO}]=0.050 M,\left[\mathrm{H}_{2}\right]=0.045 \mathrm{M},\left[\mathrm{CO}_{2}\right]=$$0.086 M,$ and $\left[\mathrm{H}_{2} \mathrm{O}\right]=0.040 \mathrm{M} .$ (a) Calculate $K_{\mathrm{c}}$ forthe reaction at $686^{\circ} \mathrm{C} .$ (b) If we add $\mathrm{CO}_{2}$ to increase its concentration to $0.50 \mathrm{mol} / \mathrm{L},$ what will the concentrations of all the gases be when equilibrium is reestablished?

$$\begin{array}{l}{\left[\mathrm{CO}_{2}\right]=0.48 \mathrm{M}} \\{\left[\mathrm{H}_{2}\right]=0.020 \mathrm{M}} \\{[\mathrm{CO}]=0.075 \mathrm{M}} \\{\left[\mathrm{H}_{2} \mathrm{O}\right]=0.065 \mathrm{M}}\end{array}$$

Chemistry 102

Chapter 14

Chemical Equilibrium

University of Central Florida

Rice University

University of Maryland - University College

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.

01:50

Consider the equilibrium f…

02:39

Calculate the equilibrium …

04:23

The equilibrium constant $…

01:54

04:50

07:35

So for the following balanced chemical equation, first thing we want to dio is find what K. C is. And this is actually very easy to Dio because in the problem, we were already given our concentrations of all the products, all the reactant at equilibrium. So all we have to do is plug it in now and we're gonna dio products times product concentration all over the concentration of our reactions, which is 0.0 for five multiplied by 0.0 86 This is equivalent to 0.5 to So this is our Casey value at our initial equilibrium. But we're told in the problem to take into account what happens if we change one of the values. What will be our final equilibrium? Constant concentrations for every substance with a change, the value of carbon dioxide. So since we're going to find new equilibrium values going to do an ice table, which means initial change and then at equilibrium, So we already know her initial concentrations we're going to use the same initial concentration are the same equilibrium concentrations that we were given in the problem itself. Except we know because we were told that carbon dioxide concentration is changing to 0.5. Everything else going to use their equilibrium concentration as the initial because we started out at equilibrium until we added more co two. So for the change, it's just excess because we have no coefficients before any of the products or reactant, that means our equilibrium at equal of room expressions for our first reactant. Her and I are site is your 0.5 minus X hydrogen 0.45 minus X current monoxide 0.5 plus X and then our water and the gases form is your 0.4 plus X. So now that we have thes equilibrium values and we know what our equilibrium constant is, we calculated it before with are given values. We can put what we have here for this last row in her ice table into an equilibrium constant expression equation so that we can solve for X that easy have 0.5 plus x times 0.4 plus x. These are products over our reactions, so you have zero point five minus x and 0.45 minus x. So if we were to simplify this and unsolved using the quadratic equation, we would find X equals 0.0 25 And since we know this X value for our new equilibrium, we can easily find our new equilibrium concentration values. You will want to do this for every substance in the reaction. Lastly, we have water. So looking at our first reacting carbon dioxide, this is the one that we had changed, that we had to do all of this four. So he no 0.5 minus X. We found the extra 0.25 and we know that this is equivalent to 0.4 eight as our equal a room concentration when it reaches equilibrium again. There, for our hydrogen, we had 0.45 minus x 0.25 We found that well 0.2 Then for carbon monoxide, our first product, we have 0.5 plus our X value. We found to be 0.25 That is equivalent to 0.0 75 water 0.4 waas our x value and we know that this equals 0.0 65

View More Answers From This Book

Find Another Textbook

01:31

Consider two bulbs containing argon (left) and oxygen (right) gases. After t…

01:11

Why do elements that have high ionization energies also have more positive e…

02:37

The equilibrium constant $\left(K_{P}\right)$ for the reaction $\mathrm{H}_{…

Consider the second-order reaction$$\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \l…

07:26

From the following heats of combustion, $$\begin{array}{c} \mathrm{CH}_{3} \…

00:59

Write chemical formulas for oxides of nitrogen with the following oxidation …

01:53

A student placed 1 g of each of three compounds A, $\mathrm{B},$ and $\mathr…

03:10

How much HCl (in moles) must be added to 1 L of a buffer solution that is $0…

08:15

Calculate the $\Delta S_{\text {soln }}^{\circ}$ for the following processes…

01:27

Which of the following does not have $\Delta H_{f}^{\circ}=0$ at $25^{\circ}…