The reaction SO₂(g) + 2 H₂S(g) ⇌ 3 S(s) + 2 H₂O(g) is the basis of a suggested method for removal of SO₂ from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (d) Would you expect the process to be more or less effective at higher temperatures?

Video transcript

Hello. Everyone in this video, we're trying to see if the reaction given over here becomes more or less effective as the temperature rises. We can find this information out by solving for our change of entropy is our delta H. Or change in entropy which is our delta S. So we're given all the values that we can use to go ahead and sell for these. So first let's go ahead and solve our delta H. So our delta H. Of the reaction is the delta age of our products minus the delta H. Of our reactant. So again we're just using the values given to us right over here in this table. So first we're being asked for the death of age of our products and we know that our products are on the right side. So we just need to match these up with this. We do that, we see that for our delta H. Of our products. Let's see here. So we have one more multiplied with zero killer jewels promote and then we're gonna go ahead and add one more Multiplied by. Let's see here we have a value of -241.82 killer joules per mole. So we see here that for our carbon here we have zero value and we have one mole of um the carbon for water, we have the delta H. Value of negative 2 41. So that's how I got this. So we're going to continue the same method for our reactant. So again we have one more multiplied by negative 110.5 units being killed jules Permal subtracting this with one mole mortified by zero killer jewels per mole. So once I put all this into my calculator I can get that my delta H. Of this reaction is equal to negative 131. killer jewels per month. Now for my delta S. The the reaction or the equation to solve for delta S. Exactly miss this except for H values will put us. So tell to us is equal to the entropy or the entropy of our reactant. So discovery. So the entropy of our products minus the entropy of a reactant. Again using the table of values given to us right over here You can see that we're dealing with this here, one mole multiplied by 5.69 jewels per mole times Calvin. And then we're adding that with let's see one mall times 188.83. Again, units being jules Permal times kelvin. And then we're gonna go ahead and subtract this with one more Times 197. Jewels Per Mole Times Kelvin. And then we're gonna go ahead and add, let's see the next reactant would be we have one more of that and then we're looking at the entropy value that's going to be 130.58 jewels per mole times kelvin. So once we put that all into our calculator, we see that the change of entropy is going to be equal to negative 133.96 jewels per kelvin. Okay, so we see here that from what we saw for their delta H value is negative. Are delta S value is also negative. So when we have both values being negative, this reaction will only be spontaneous at lower temperatures. So of course as the temperature rises, then the reaction will be less effective. Let's go ahead, write that answer out. So we can say that the reaction will be less effective as the temperature prices. Alright, so this is going to be my final answer for this problem. Thank you all so much for watching.