Without doing any calculations, determine the signs of ΔS_sys 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. c. 2 N₂(g) + O₂(g) → 2 N₂O(g) ΔH°_rxn = +163.2 kJ

Video transcript

Welcome back everyone in this example, we need to identify the sign of change in entropy of our system as well as the change in entropy of our surrounding for the reaction below without doing calculations and identify the temperature condition in which the reaction would be spontaneous. So we're given the below reaction where we have two moles of hydrogen gas reacting with one mole of chlorine gas to produce two moles of hydrogen chlorine gas. And we're given our entropy of our reaction as the value negative 185 kg joules now because we can't do any calculations here. We want to pay attention to the moles of gas on each side of our equation. And that is due to the fact that this question is asking us about change and entropy and we want to recall that our entropy change is going to be a measure of the random redistribution of thermal energy of a system. And because it's the random redistribution measure of thermal energy, we're going to take note of how our gasses are exchanged in this reaction. So we should recognize that on our product side we have a total of three moles of gas so of gaseous products, sorry reactant. But then on our product side we just have two moles of our gaseous products. And so we went from having a higher amount of gaseous re agents to a lower amount of gaseous re agents. And so this distribution of thermal energy here of our gaseous re agents is a negative change because we ended up with less less gas produced on the product side. And so we should understand that this is going to relate to a decrease in our measure of entropy I'm sorry this says entropy And so because we have a decrease in entropy of our system here we would say that the change in entropy of our system which is our reaction here is going to be negative. So this is going to be our first answer. And now we want to figure out the entropy change of our surroundings. Now. To figure that out. We want to recall the below formula where our entropy change of our surroundings is related to our entropy change of our system divided by the temperature that our system is within And according to our given reaction we have a negative entropy or sorry entropy change of our reaction. It's a value being negative. 185 kg joules so that's why it's negative in our formula here. So let's underline this value. Now based on this formula we've recalled we can see that our entropy change of our surroundings is in the numerator just like our entropy change of our system is also in the numerator. And so because they're both the numerator we can say that our entropy our sorry our entropy change of our surroundings is directly related to our entropy change of our system. And so we would say that therefore based on this direct relationship because we understand that our Entropy of our reaction is negative or less than zero. We can say that our entropy change of our surroundings which is directly related to our entropy is going to also be negative or we can say less than zero as well. So this is going to be our next answer. And lastly according to our prompt, we need to determine the temperature condition in which our reaction is going to be spontaneous. So we want to recall that when it comes to having a spontaneous reaction that means that we want to have a free energy change which we call is represented by delta G of our reaction that is going to be negative and we want to recall that This only applies at low temperatures and we can relate our change in free energy to entropy and to entropy by recalling that our change in free energy for a spontaneous reaction which is negative is going to equal our change in entropy which is also negative according to our given prompt subtracted from our temperature multiplied by our change in entropy which is also as we determined negative And this should be for temperature low temperature because as we stated according to the prompt, the reaction is going to be spontaneous and that means that our change in gibbs free energy should be negative only at low temperatures. So low temperatures would be the condition for our reaction to be spontaneous. So everything highlighted in yellow is going to be our final answer to complete this example. I hope that everything I reviewed was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.