Consider the reaction 3 CH41g2¡C3H81g2 + 2 H21g2.
(a) Using data ...

Question

Consider the reaction 3 CH41g2¡C3H81g2 + 2 H21g2.
(a) Using data from Appendix C, calculate ΔG° at 298 K.

Accepted Answer

Welcome back everyone in this example. We're told that at 2 98 kelvin, what is our change in standard free energy for the reaction where we have magnesium carbonate solid decomposing to form magnesium oxide, solid and one roll of carbon dioxide gas. So as we stated, this is a decomposition reaction and we have to determine what's going to happen to the magnitude of our standard free energy of our reaction if our temperature is decreased. So we have a two part question and we're going to go ahead and begin with solving for part one where we want to recall to find our change in standard free energy of our reaction. We're going to take our change in standard free energy of our products. And we want to subtract that from the change of standard free energy of our reactant. So we're going to designate the reactant by the color red here. And we would find the free energy change of the products and reactant from our textbooks or online. So, plugging in what we know to our formula, we can say that our free energy change of our reaction is equal to first Beginning with the free energy change of our products. We want to begin with our two products we have as we read one mole of our magnesium oxide solid, Which is multiplied by its free energy change according to our text books. As the value negative 569.3 with units of Kayla Jewels per this is then added to the second product where we have one mole of CO. Two from our reaction multiplied by carbon dioxide standard free energy change according to our text books. As to value negative 394.4 kg joules per mole. So this completes the some of the standard free energy change of our products. And now we're going to subtract and read from the some of the standard free energy change of our reactant. We just have our one reactant which is our one mole of our magnesium carbonate which is multiplied by its centered Free energy change according to our text books. As the value negative 1112 or sorry, negative 1012.1 units of kilo jewels per mole. So now what we want to do is simplify this in our calculators. So what we're going to get is that our Free energy change of our reaction is equal to for the sum of the free energy change of our products. We should get the value negative 963.7 kg jewels Permal. So this completes the some of the standard free energy of our products which is subtracted from the standard free energy. Some of our reactant where we should get the value negative 1012.1 kg jewels per mole. And so simplifying this, we would say that our free energy change of our reaction is equal to A value of 48.4 kg joules per mole. And this is because we recognize that we're subtracting a negative so we're really adding here. So we should recognize that this is a positive value. We have a positive magnitude right now for our free energy change of our reaction. And this means that this reaction is going to be non spontaneous. So we should recall that that means that our reactant doesn't have enough energy to start up the reaction. So we need energy to be added to our react inside in order to get our reaction going. And this would make sense. Especially if we take note of our number of gaseous re agents in our reaction. And we can see that we only have a solid reacting here, but on our product side we have one mole of our gaseous Reactant c. 0 2. So this will lead us into figuring out part two of our prompt where we're going to determine the magnitude of our free energy change of our reaction if the temperature is decreased and how this leads us to answering this is that because we recognize we have one mole of our gaseous products produced on the product side. We have an increase in our moles of gas in our reaction. And so we would recall the relationship where our change in our free energy of our reaction is equal to our change in entropy of our reaction delta H, which is then subtracted from our temperature of our reaction which is multiplied by the change in entropy of our system here. Our system being our reaction now from this formula, we want to recall that entropy is going to be our measure of our redistribution of thermal energy in our system. So being our reaction. And so that is going to relate to our gas or the gas is involved in the reaction. And we can see that we as we stated, have a increase in gas on our product side of our reaction where we have one mole of our gaseous co two meaning that we have a positive change in entropy. And so because we recognize according to this formula that we have recalled that all of our variables being our change in free energy are entropy and our change in entropy are all in the numerator. So they're all directly related to one another as well as directly related to temperature. We can say that because our specifically free energy of our reaction is directly related to temperature. Therefore, as temperature increases our change and free energy of our reaction is also going to increase or become positive in magnitude. And so to complete this example, we would confirm that our final answers are going to be our calculated free energy change of our reaction which is highlighted in yellow here as well as the fact that our magnitude is going to be positive for our free energy of our reaction as temperature is increased. So I hope that everything I reviewed was clear this is going to correspond to choice. Seeing the multiple choice. If you have any questions, leave them down below, and I'll see everyone in the next practice video.

Consider the reaction 3 CH41g2¡C3H81g2 + 2 H21g2. (a) Using data from Appendix C, calculate ΔG° at 298 K.

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