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Problem 38

Find $\Delta S_{\mathrm{rxn}}^{\delta}$ for the combustion of ammonia to nitrogen dioxide and water vapor. Is the sign of $\Delta S_{\mathrm{rxn}}^{\circ}$ as expected?

Answer

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## Discussion

## Video Transcript

want to determine the overall change and standard more entropy for the combustion of ammonia into enoh to gas and water vapor. We need to begin by writing out the combustion reaction and balancing it. So after we write out the appropriate reactions and products, we can balance this equation in the following way. If we start by looking at the hydrogen, we see that we would need a coefficient of to here and three here in order to balance him out. However, this would lead to an odd number of oxygen's from water and lied to us having to use fractions or the coefficients. So if we were to multiply that by two to make that oxygen from water an even number and we would have coefficient of four here for ammonia and therefore have four nitrogen ins on both sides, and that would lead to 12 hydrogen ins on the reactive side so that we would need a six or the coefficient of water have 12. Hydrogen is on each side, and now we have eight Oxygen's from N 02 On the products was six oxygen's from H 20 which comes out to a total of 14 oxygen atoms. So we need a coefficient of seven for 02 to balance those out. So now we need to come up with a prediction. For what? We think the sign will be positive or negative for Delta s reaction. In order to do this, we examine all the species in the reaction that are gasses. Since gases have the greatest standard Miller entropy out of the three states of matter we look, we can see that all of the chemical species involved in this reaction are already in the gas estate. So we need to count the total number of moles on the reactant side and product side on the react. Inside, we see that we have four plus seven. So 11 moles of gas and we're forming or plus six. We're 10 moles of gas. So the total number of gas molecules decreased from 11 to 10 as a result of the reaction. And since gases have the greatest standard bowler entropy, if we have more moles of gas than we have a greater amount of entropy on the reactant side and therefore more disorder due to the random movement of those gas particles than we do on the product side. So that compared it to the reactant. The products are more ordered and have a lower standard Mueller entropy. And so we can predict that the change and Stayner more entropy is negative so that we can decrease the number of moles of gas from 11 down to 10. And now we use this equation too. Solve out for a value for Dell tests of reaction where we take the summation of the number of moles of each chemical species multiplied by that chemical species standard molar entropy value. And then we do the same for the reactant side so that we have the total standard Miller entropy of the products minus the total standard more entropy of the reactant. We break up the terms into the products in the reactant and we can start on the product side of this chemical reaction that we balanced and we see that we have four moles of n 02 and six moles of H 20 formals of n 02 and six moles of H 20 We can look up the values for the standard molar entropy, ease of each one of these chemical species in the appendix. And when we cancel out moles, we're left with units of jewels per Kelvin which corresponds to units of total standard Moeller entropy. When we do the math, we get this value for the total standard Miller entropy of the products. And now we do the same for the reactant. Since we have formals of NH three and seven moles of 02 formals of NH three and seven moles of 02 we look up the standard more entropy values for each and H three n and 02 in the appendix and we come out to this value for the total standard Miller entropy of the reactions And now we have we have to subtract u minus are we have our values for P and are And when we do that, we get this value and this is this corresponds to the change in state or more entropy, pretty combustion of four moles of ammonia gas. If we were to define the change in standard more entropy of this reaction for the combustion of one mole of NH three gas than we would just divide this answer by four to account for the formals that we used to balance the reaction, which comes out to negative 28 0.8 jewels per kelvin. Anyway, we went to you right out the final answer. In each case, we have a negative sign and that matches what our prediction was. So we can see that, Yes. The sign of Delta s reaction matches our expectations.

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