Enroll in one of our FREE online STEM summer camps. Space is limited so join now!View Summer Courses

Problem 57

For the gaseous reaction of xenon and fluorine to…

02:51

Need more help? Fill out this quick form to get professional live tutoring.

Get live tutoring
Problem 56

Consider the combustion of butane gas:
$$\mathrm{C}_{4} \mathrm{H}_{10}(g)+\frac{13}{2} \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g)+5 \mathrm{H}_{2} \mathrm{O}(g)$$
(a) Predict the signs of $\Delta S^{\circ}$ and $\Delta H^{\circ} .$ Explain.
(b) Calculate $\Delta G^{\circ}$ by two different methods.

Answer
Check back soon!

Topics


Discussion

You must be signed in to discuss.

Video Transcript

we're given the combustion reaction for one mole of butane reacting with oxygen to produce carbon dioxide and water. And this equation is already balanced or the combustion of one mole of butane in part A. We want to determine the signs for you change an entropy and the change in entropy of the system meaning that will they be positive or negative to determine the change in entropy Delta s meaning to examine the total number of gas molecules on either side of the reaction, We look at the reaction side that we see that we have a total of one was 6.5. So 7.5 moles of gas going to four plus five or nine moles of gas. And so that means that the total number of moles of gas increases as a result of the reaction and gas is compared to solids and liquids have the highest amount of disorder in there and they're particle movement due to the random motion of those those gas particles and therefore they have a greater amount of entropy. So do you increase the number of gas molecules means that we increase the entropy in order to increase the entropy. The change has to be positive or greater than zero. So Delta s, we should expect should have a positive change. And now, for the the change in an entropy of the system, we are told that this is a combustion reaction and all combustion reactions give off heat and therefore that heat energy and be written as a product of the combustion reaction. And because of that, the system gives off heat energy to the surrounding sue. The system loses that entropy change, and so therefore, the entropy change of the reaction we should expect should have a negative sign. Since this is an extra thermic reaction on purpose e, we want to find the overall change and Gibbs free energy at standard conditions for this reaction, using two different methods first method that we're going to use to find this is the the general equation for directly solving for this, based on the change in Gibbs free energy of formation at standard conditions, where the products multiplied by the number of moles of each species in the reaction and adding all of them for the products and then subtracting the same quantity of for the reactive and so we can use this equation to directly saw for Delta G. We start with the product side. We see that we have four moles of CO two and five moles of H 20 So four moles of CO two and five moles of H 20 and we can look up in the appendix B Delta G of formation values for co two and H 20 at standard conditions. When we multiply those with the total number of moles of each species participating in this combustion of one mole u tane we see that the units of moles cancel out to yield units of killing JAL's or overall energy change of the products. It comes out to this value after we do the math and now for the reactant. So we see that we have one more of butane in 13 halves of a mold of oxygen. So we include that one mole of butane. But remember that for they'll teach values that the the end for the change in Gibbs Free energy values Delta G and Delta H that elements that exist in their standard states, such as molecular oxygen, have a delta h of formation value of zero. So there is no free energy change Delta G value associated with the formation of 02 And so it has a value of zero, and we therefore do not have to include it in the Gibbs free energy change of the reactant. So the only change in energy from the reactant is is factory in by the U tane. So we look up the value for the delta G of formation for butane. I want to play it with that one mole, that of it that is combusting to get the total changing Gibbs free energy of the reactant. And then again consulting this equation, we do the Gibbs free energy change of the products, minus that of the reactions. So this is Delta G for the products. This is Delta G for the reactant, and we subtract those two, we get this answer for Delta G of reaction. The second method that we're going to use to solve for this variable is to use a very similar equation to this one. But instead for finding Delta H and again Delta s. And this is the formula written out for finding Delta each of reaction and for Delta s U Industry police. All of the h is with with s is for the entropy. And after we do that, we want to plug it into this equation. Delta G equals Delta H minus t Delta s because we can solve for Delta H in Delta s using those equations. And we know that the temperature of standard conditions is 298 Kelvin and so we can sell for Delta G. So starting with Delta H begin, do the total entropy change of the products minus the total entropy Change of the reactant. Starting with the products we again have formals of co 25 moles of H 20 Formals Co two and five moles of H 20 each multiplied by their well to each of formation values then added together, gives us the total well to each of the products. And then, for the reactant, we have one mole of you teen in 6.5 moles of oxygen ensue One more butane But again, just like Delta G, oxygen exists naturally as a die atomic gas and therefore has a delta H of formation value of zero. So the total entropy change of the reactant. It's just is just the total entropy change of that one more of beauty. And again we do products minus reactant to get delta h of reaction. And now we do the same for finding the entropy. We have the same striking metric coefficients of four moles of CO two and five moles of H 20 And this time in the appendix we look up the standard molar entropy values for each one of these compounds. And when we cancel units, we get, we can't solve the most. We get units of Jules per killed into correspond to total entropy of the products. And then we do the same for the reactant. And this time we can include the 6.5 moles of oxygen. Since it has a non zero entropy value. And when we do that math, we can get the total entropy change of the reactant. And just like before, we do the total entropy of the products minus a total entropy of the reactant. You get Delta s of the reaction and we need to divide this by 1000 to convert it into units of Hillary jewels. per Kelvin instead of jewels per Kelvin. Because when we plug in the values that we found into this delta G of reaction equation, we need to have similar units in order to be able to subtract the energy values. So again, the Delta, each of reaction value that we found was negative. 2657.13 Killah JAL's and so that goes here. The temperature is 298 Kelvin at standard conditions and we just found Delta s to be 155.9 Jules per Kelvin. But again, when we plug it into this equation we needed after we can't solve the units of Kelvin, we need to have the same energy units, which is why we had to divide by 1000 for the entropy value. And now, when we plugged at all in, we can get a delta G of reaction equal to negative 2000 and 3.6 killer jewels, which is very close to the Delta G that we found using the other method

Recommended Questions