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At $1024^{\circ} \mathrm{C},$ the pressure of oxygen gas from the decomposition of copper(II) oxide (CuO) is 0.49 atm:$$4 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g)$$(a) What is $K_{P}$ for the reaction? (b) Calculate the fraction of CuO that will decompose if 0.16 mole of it is placed in a 2.0 -L flask at $1024^{\circ} \mathrm{C}$. (c) What would the fraction be if a 1.0 mole sample of $\mathrm{CuO}$ were used? (d) What is the smallest amount of $\mathrm{CuO}$ (in moles) that would establish the equilibrium?

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(a) $K_{p}=0.49$(b) 0.23(c) 0.037(d) $0.037 \mathrm{mol} \mathrm{CuO}$

Chemistry 102

Chapter 14

Chemical Equilibrium

Rice University

University of Maryland - University College

University of Toronto

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.

06:57

At $1024^{\circ} \mathrm{C…

02:53

Be sure to answer all part…

01:49

At $1000^{\circ} \mathrm{C…

04:46

A stream of gas containing…

03:23

The equilibrium constant f…

02:49

Consider the decomposition…

01:34

When one mole of carbon di…

01:47

An equilibrium mixture for…

02:00

$\mathrm{Kp}$ is equal to …

So for this problem, we're looking at the falling balanced chemical equation and we want to first find what KP is, and we can find this fairly easily. So talk about KP misses that equilibrium constant in regards to partial pressure and that the only things that make up the KP equation are substances in the gaseous state. And we only have one here, so that means KP is on Lee made up of the partial pressure of auction, and we were given that in the problem itself. So we know KP is zero point four nine. So now that we know KP weaken, turn our focus to the next component we want to look at and then is the percent of our reactant that has decomposed. So first we want to do is find how many moles of auction we have, since this is a given value and we can use it to find further information that we need that we can find out about our reacting because this is one of our products and we want to know about decomposition. I want to really react. It decomposes, okay, produces the products here. So first we want to find moles of auction. So we're going to rearrange the ideal gas equation. It's that it looks as follows and then we already know all these values his Bieber given them in the problem itself. This is two leaders with volume of the ideal gas constant 0.0 8 to 1. The temperature in Calvin is 12 97 And if you were to calculate this out, you had fund that for the number of moles. We have 9.2 times 10 native, three bowls. And since we know the number of moles auction, we can use thestreet geometry of the balanced chemical equation itself to find the moles of our reacted. So, in order to do so, you would take the moles of auction that we found. And we know now for every one more of oxygen, we have four moles off our reactant based on the story geometry here so we can find that this is equivalent to zero point zero 37 more 3.7 pies, 10 to the negative, too moles of our reactant. So, in order to find the percent decomposition from this, we can go ahead and take what we found to be the moles at equal room and invited by what we were told was the initial moles of our reacted and we used the initial moles because we're looking at how much has decomposed at the equilibrium value and this is equivalent to 0.23 in order to find the percent would multiply this by 100 find that 23% had decomposed. So we're gonna go down all of its do the next parts next, its focus on C. So we know that even if we change the initial moles auction is the only gaseous species involved in equilibrium, constants equal of room, cause it's not gonna change. And in order to find the new percent for see, we can do just like we did before 3.7 times 10 too negative, too, because they're auction value is not changing, and only in the initial is. And we were told, we're using 0.0 moles for this hypothetical part in C, which is equivalent to 0.37 who multiply this by 100. We find 3.7% he composed for our reactant. So looking at D, if we want to know the minimum amount of our reaction that we need to make the create correct amount of oxygen. We're going to continue using the same KP value. And since we already calculated it's above, we know that 0.37 moles off the reactant is going to be the minimum needed for us to reach that equilibrium value that we saw at the KP or the equilibrium constant for oxygen.

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