Enroll in one of our FREE online STEM bootcamps. Join today and start acing your classes!View Bootcamps

Ohio State University

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72
Problem 73
Problem 74
Problem 75
Problem 76
Problem 77
Problem 78
Problem 79
Problem 80
Problem 81
Problem 82
Problem 83
Problem 84
Problem 85
Problem 86
Problem 87
Problem 88
Problem 89
Problem 90
Problem 91
Problem 92
Problem 93
Problem 94
Problem 95
Problem 96
Problem 97
Problem 98
Problem 99
Problem 100
Problem 101
Problem 102
Problem 103
Problem 104
Problem 105
Problem 106
Problem 107
Problem 108
Problem 109
Problem 110
Problem 111

Problem 42

(a) Find $\Delta S_{\text { rxn }}^{\circ}$ for the formation of $\mathrm{N}_{2} \mathrm{O}(g)$ from its elements.

(b) Calculate $\Delta S_{\text { univ }},$ and state whether the reaction is spontaneous

at 298 $\mathrm{K}$ .

Answer

Check back soon!

You must be logged in to like a video.

You must be logged in to bookmark a video.

## Discussion

## Video Transcript

we need to determine the total change in entropy. Delta s of reaction for the formation of into, oh gas from its elements. We further than need to calculate the built s of the universe as a result of this reaction and determine whether this reaction proceeds spontaneously or non spontaneously. We begin by writing out the formation reaction of into a gas from its naturally occurring elements of nitrogen and oxygen. We know that the or the delta H of formation pour into a no to gas zero. And so it does not take any energy to produce either one of those by atomic gases since they occur naturally so that those are what we use in order to form and to Oh gas. We need to balance this reaction so that we have to one in two. So when we when we count up, we so we have four nitrogen atoms on each side and two oxygen atoms on each side. And now we use this equation to determine Delta s of reaction. By doing the total entropy of the products minus a total entropy of the reactant, we can break up the products in reactant Sturm's and two p and R in starting on the products, said, We see that we only have one product and 20 gas. We have two moles of that product Foreman. So two moles of n 20 we look up. It's standard molar entropy value in the appendix. Multiply those two together and we have the total entropy of the products. Now we do the same. For the reactant, we have two moles of n two and one mole of 02 two moles Event to in one mole of 02 and those each have a standard miller entropy value that we can look up and again. As we do the math, we cancel out the units of moles to get total Moeller to get the total entropy of the reactant sin units of jewels per Kelvin. And now we do. P minus are and we have values for P and R. And when we subtract them, we should get Delta s of reaction to be equal to this value. However, this is for the production of two moles of n 20 gas and four formacion reactions way we want to you know, the values for Delta s Delta h etcetera for the production of one mole of that desired compound. And so, since this is the delta s of reaction for the production of two moles, we should divide it by two to get the delta s of reaction for the production of one mole of into a gas. And that comes out to about negative 74 0.3 jewels per kelvin. And that is the answer part A for the delta s of reaction. And in per p again, we're solving for Delta s universe to help us determine whether this reaction is spontaneous or not. This is the equation that that we use. And we have another equation to find Delta G of reaction at standard conditions which is identical to the one used to find Delta s of reaction. We have Delta G that standard conditions of formation for an indie appendix for different chemical species and four information reactions. Just like the values for Delta H Delta G values will be zero. Since these species occur naturally, information reactions were forming one mole of product from its naturally occurring elements. And so there's naturally occurring elements will always be on the reactant side will always have a Delta G and delta each value of zero since no energy is required to form them. So we just examine the product side. Can we have two moles of into a gas through two moles of n 20 multiplied by its standard change in Gibbs free energy. And we get the total change in Gibbs free energy since the reactors term is zero to be this value. And now we can rearrange this equation for Delta s universe. And Saul, for it in this way equals delta G of the reaction at standard conditions divided by the negative temperature which we're told is 298 Kelvin. So we should convert this value for the Delta G of reaction into jewels by multiplying it by 1000 so that we have 208 0.4 times 10 to the third Jules, divided by negative 298 killed in. When we do that division, we should get an answer of about negative 600 99.3 Jules Per Kelvin. But again, this is for the formacion of two moles of n 20 ensue for the formation of one mole of into A. We would divide this by two to get about an answer for Delta s universe to be about negative 300 49 0.7 jewels per kelvin. Now, in order to determine whether this reaction of spontaneous or not you look at the sign of Delta s universe since its negative, That means this reaction is is non spontaneous. If that value were positive than we would see that the reaction is spontaneous, We can also consider the sign of Delta G. See that since it is a positive value, that means it is non spontaneous. If Delta G were negative, it would be spontaneous. So Delta G is opposite of Delta s universe in that way. So we can say that this reaction is non spontaneous at 298 killed him.