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Problem 37

Find $\Delta S_{\text { rxn }}^{\circ}$ for the r…

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Problem 36

Find $\Delta S_{\text { rxn }}^{\circ}$ for the combustion of methane to carbon dioxide and liquid water. Is the sign of $\Delta S_{\text { rxn }}^{\circ}$ as expected?

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in this problem. We want to determine the change in standard Moeller entropy as a result of the combustion of methane gas with oxygen in order to produce carbon dioxide, gas and liquid water. The first step that we have to take in this problem is to balance the combustion reaction after we write it out this way involving the species that are told to us in the problem statement. We start with the methane and we combust one mole of it. We see that on the right side one urban Adam from carbon dioxide balances out that mole of methane so that we have one carbon atom on each side of this reaction. Furthermore, we see that that single mole of methane produces for hydrogen atoms on the reactant side. So we need to strike you metric coefficient of two for water in order to balance out the hydrogen. And now, finally looking at the product side, we see that we have two oxygen atoms from co two was two oxygen atoms from water for a total of four oxygen atoms on the product side. And so we need a stroke, you metric coefficient of 240 to order to balance that out. So that is the balanced combustion reaction that we're working with in this problem. And now we need to use this equation in order to calculate early in order to calculate the change in standard Mueller entropy of the combustion reaction. I first adding up all of the standard Mueller entropy. Ease of each chemical species in the products and multiplying each one by the number of moles involved in that reaction. To get the total standard molar entropy of the products and then subtract the total standard miller entropy of the reactive. So I have isolated these terms. Azaz products P and reactant are so we can look at each one individually products in reactant starting on the product side, you see that we have one more of co two in two moles of H 20 So that's one more co two in two moles of H 20 Then we look up in the appendix what the values are the correspond to those chemical species where their standard Moeller n trapeze and then the reason that we have to multiply it by the number of moles given by the striking metrical efficient in the balanced reaction Sue that when we multiply the values together, units of moles cancel in each case so that we are left with total more entropy units of jewels per kelvin. So when we right that all out and do the math for the total standard molar entropy of the products, we get this value and now we do the same for the reactive. And when we look at the reaction, we see that we have one more of methane and two moles of oxygen. So one more methane in two moles of oxygen and we multiply them with each one of their standard Moeller and herpes. And then we add them together because of the summation in the equation, we get this value and now from the equation we do e minus are. And this is our value for P. This is our value for our and we perform that subtraction and this is the answer that we get. And the other part of this problem asks us if the sign of this answer is what we expected. We see that we have a negative sign and if we go back to the reaction that were given and we consider the change in an entropy of this reaction. We need to consider the gashes species in this reaction. Remember that entropy is the amount of disorder in a given system and also recall that were any substance. That substance is in a gaseous state, compared to a solid or liquid state than the motion of it's particles is random and therefore more disordered compared to a solid or liquid state. Were they were. The particles are more organized and have less freedom of movement. And so the effect of liquid or gaseous or liquid or solid states would be negligible compared to gas estates, because gas is again produce the most disorder and therefore the greatest amount of entropy in a given system. So we have to compare the total number of moles of gashes, species before and after the reaction we look, we see that we have a total of one plus two, 23 moles of gas before the reaction on the reactant side, and we form one mole of a gas on the product side, and the change in an entropy is the products minus the reactive in one minus three comes out to a negative value. Remember that that gases or them have the greatest, the greatest total standard Mueller entropy compared to the other two states of matter solids and liquids. And so if we have three moles of of gas compared to one mole, then we have mawr, a greater amount of of disorder on the react inside compared to the product side, so that as we make that change from the reactions to the products, the system becomes more ordered as result of having fewer gaseous molecules and therefore the entropy increases. And so we we should predict that the delta s of reaction would therefore have a negative sign you choosing negative change in entropy of the system is result of that change in gashes species. And again we see that negative value and the answer that we calculated So weaken say that yes, the sign that that we got matches, what we predict

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