Skip to main content
Ch. 18 - Free Energy and Thermodynamics

Chapter 18, Problem 41a

Given the values of ΔH°rxn, ΔS°rxn, and T, determine ΔSuniv and predict whether or not each reaction is spontaneous. (Assume that all reactants and products are in their standard states.) a. ΔH°rxn = +115 kJ; ΔS°rxn = -263 J/K; T = 298 K

Verified Solution
Video duration:
4m
This video solution was recommended by our tutors as helpful for the problem above.
1477
views
1
comments
Was this helpful?

Video transcript

Hello everybody. So in this video we're gonna go ahead and calculate for our delta as of the universe as well as predict the reaction is going to be spontaneous or not spontaneous. And of course we're given this data right over here to help us. So let's first recognize the equation for our delta us of the universe. So delta sf universe is equal to the delta S of our system, plus the delta us of our surroundings. Let's first calculate for the delta S of the system. So the first component of our equation. So the delta S of the system is actually equal to the delta S. Of direction. And conveniently we are given that in this problem right over here. So then our delta as of the reaction is equal to negative 1:54 jewels per kelvin. Now, solving for the delta S of the surroundings. Next that is equal to negative delta H. Of a reaction divided by T. For temperature. So again we have those given to us right over here and right over here. So let's go ahead and plug those values in. Of course a negative and negative will cancel to a positive. So we have a positive 256 killer jewels. We want our units to match in this equation. We have jewels here. So let's convert this killer jewel into jewels. So 103 jewels goes on top and one killer jewel on the bottom. As you can see the killer jewel unit will cancel nicely. All right. And of course all of this is going to be divided by R. T. for temperature. So 3 15 cupboards. So my numerical value will come out to be 812.69 units will be joules per kelvin. So now they're both components of this equation. Right over here, let's go ahead and plug those numerical values in them. So our delta S of the universe is equal to the delta S of the system. So we have negative 1 50 for jewels per kelvin. And then we're gonna go ahead and add that with our delta S of the surroundings, which we have solved for to be 812.6 nine jewels per kelvin. In that case our delta s of the universe will be equal to 659 jewels per kelvin. That's the first part of our answer. Second part is if the reaction is going to be spontaneous or not. So how do you see if the reaction will be spontaneous is by calculating for our delta G, delta G is equal to delta H minus T. Multiplied by our delta S. Of course we have solved for these values and we're also given these values. So let's go ahead and plug them in. So delta G is equal to This negative 2 56 killer jewels minus R T. So 3 15 kelvin's time limps are adult us, which is negative 1 50 for jules per kelvin. Again, we want all of our units too much. We have killed jules here, let's convert this jewels into jewels, So we'll have one killed jewel on top and 10 to the third jewels on the bottom. As you can see, then our jewels will cancel out. And the kelvin's will also cancel out here. Alright, so plugging that into my calculator. My numerical value for Delta Jeep is -207. My unit is going to be just killed jules. So are a delta G value you see is clearly a negative number. Whenever we have a negative delta G, this just means that reaction will be spontaneous. Alright, so this is going to be the second part of our answer. And again we have that delta s of our universe is positive 659 joules per kelvin. And the reaction is going to be spontaneous. Thank you all so much for watching.
Related Practice
Textbook Question

Without doing any calculations, determine the signs of ΔSsys and ΔS surr for each chemical reaction. In addition, predict under what temperatures (all temperatures, low temperatures, or high temperatures), if any, the reaction is spontaneous. a. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH°rxn = -2044 kJ

2689
views
1
comments
Textbook Question

Without doing any calculations, determine the signs of ΔSsys and ΔSsurr for each chemical reaction. In addition, predict under what temperatures (all temperatures, low temperatures, or high temperatures), if any, the reaction is spontaneous. c. 2 N2(g) + O2(g) → 2 N2O(g) ΔH°rxn = +163.2 kJ

596
views
Textbook Question

Calculate ΔS surr at the indicated temperature for each reaction. d. ΔHrxn ° = +114 kJ; 77 K

1273
views
Textbook Question

Given the values of ΔH°rxn, ΔS°rxn, and T, determine ΔSuniv and predict whether or not each reaction is spontaneous. (Assume that all reactants and products are in their standard states.) c. ΔH°rxn = -115 kJ; ΔS°rxn = -263 J>K; T = 298 K

1060
views
Textbook Question

Given the values of ΔH°rxn, ΔS°rxn, and T, determine ΔSuniv and predict whether or not each reaction is spontaneous. (Assume that all reactants and products are in their standard states.) a. ΔH°rxn = -95 kJ; ΔS°rxn = -157 J/K; T = 298 K

1500
views
Textbook Question

Given the values of ΔH°rxn, ΔS°rxn, and T, determine ΔSuniv and predict whether or not each reaction is spontaneous. (Assume that all reactants and products are in their standard states.) c. ΔH°rxn = +95 kJ; ΔS°rxn = -157 J/K; T = 298 K

277
views