Without doing any calculations, determine the signs of ΔSsys
and ...

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

Accepted Answer

Welcome back everyone in this example, we need to identify the sign of our change in entropy of our system and change in entropy of our surroundings for the below reaction without doing calculations and identify which temperature condition where the reaction would be considered spontaneous. So based on our below reaction, we should recognize that since we have oxygen as a reactant and our products being carbon dioxide in water, this is a combustion reaction where we have two moles of benzene reacting with 15 moles of oxygen in gaseous form to produce 12 moles of carbon dioxide gas and six moles of liquid water. Were given our entropy of our reaction as a negative value being negative, 6534 kg joules. Now, because part one of this question is focused on our symbol for entropy of both our system and the surroundings. We want to recall that change in entropy is going to be the measure of random thermal redistribution. Or we can say of random redistribution of thermal energy of a system. And so in this case, since this is focused on the random redistribution of thermal energy, we're going to focus on how our gas gas is involved in the reaction are Exchanged in the reaction. And so we can see that on our reactant side we went from moles of gaseous oxygen to on the product side, a lower amount of moles being 12 moles of carbon dioxide gas. And so we had what we can say as a negative change in moles of gas since we went from a higher amount to a lower amount and this would correlate to a decrease in entropy. So we would say that entropy is going to be a negative change here. And this is specifically in reference to our system which is going to be our reaction. Now. With regard to our change in entropy of our surrounding. We want to recall the following formula where our change in entropy of our surrounding is equal to our change in our entropy of our reaction of our system because our reaction is our system which is divided by temperature. And so we should recognize from the prompt that our entropy of our reaction is given as a negative value here, the negative 6534 kg joules. So we're going to plug that in as negative in our recalled equation below. And because we can see that our entropy change of our system is less than zero because it's negative. We would understand that since it's in the numerator just like our entropy or sorry entropy of our surroundings is in the numerator, it's going to be directly related to each other. And so that means that if our entropy of our system is negative or less than zero, we would say that therefore our entropy of our surroundings are also going to be less than zero or negative. And so therefore we would say we can just write that our entropy of our surroundings is negative. So these mean the same thing and as we stated, that is due to their direct relationship to one another in our equation here that we've recalled. And so so far we can highlight two answers. We have determined that our entropy of our system should be negative based on a decrease in our moles of gas in the reaction. And our entropy of our surroundings should also be negative because we have a negative entropy of our reaction which is directly related to the entropy of our surroundings. So the next part of our prompt says to identify in which temperature condition the reaction would be spontaneous. So we want to recall that for a spontaneous reaction we should have a change in free energy of our reaction that is going to be negative. And if we recall the following formula where our change in free energy of our reaction can be found from taking the entropy of our reaction subtracted from the temperature multiplied by the change in entropy. We should recognize that if our free energy of our reaction is negative, then that means that our entropy of our reaction is negative and so is our entropy of our reaction, which means that our temperature should also be a low value. And so we can say that our free energy of reaction is only negative at low temperatures only meaning we can only have a spontaneous reaction if our temperature of our reaction is low. And so for our third final answer, we would highlight that our temperature should be low for our reaction to be spontaneous. So everything highlighted in yellow represents our final answer, which corresponds to choice C. In the multiple choice. I hope everything. Our view was clear. If you have any questions, leads them down below, and I'll see everyone in the next practice video.

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