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Problem 62

The U.S. government requires automobile fuels to contain a renewable component. Fermentation of glucose from corn yields ethanol, which is added to gasoline to fulfill this requirement:

$$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g)$$

Calculate $\Delta H^{\circ}, \Delta S^{\circ},$ and $\Delta G^{\circ}$ for the reaction at $25^{\circ} \mathrm{C}$ . Is the spontaneity of this reaction dependent on $T ?$ Explain.

Answer

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## Discussion

## Video Transcript

were given a reaction and we went to find the values for Delta H l s in Delta G for this reaction. This is the formula that we can use to find the change in m pulpy built to each of the reaction. We first find the total entropy change of the products using the stair, the standard entropy, information values of each chemical species and then subtract the total entropy change of the reactions in a similar way to find the change in entropy. Delta s of the reaction We find the total entropy of the products using the standard Mueller entropy values of each species and subtract the total entropy of the reactive. We have two different options for finding the changing Gibbs free energy of the reaction. Delta G. We can either use the above equation which has a similar structure, for we're finding Delta H in Delta s where we find the total change Gibbs free energy of the products minus that of the reactant using the gives free energy change of formation values at standard conditions. In each case, we could also use the below equation to find Delta G from the values of Delta H and Delta s that we found previously and plugging in the temperature of 298 health in. So starting with Delta H, we first need to find the total entropy change of the products. And on the product side, we see we have two moles of ethanol in two moles of CO two So two moles of ethanol in two moles of CO two and that corresponds to the end in this equation. And now we look up the standard mental p of formation values for each one of those species so that we multiply it by each one of those Moeller values that the units of moles cancel could give total energy units and kill it, Jules. And that is the total entropy change of the products for this reaction. And for the reactant side, we see we just have one mole of glucose and that single malt glucose has its own standard and tilapia formation value, and that when multiplied with that single mole, we have the the total change in entropy of the reactant. And now we subtract the total entropy of the products minus the reactant. You get our final value for Delta h of reaction to be about negative 69 killing jewels. And now we use this equation in the middle for finding Delta s of reaction The change in entropy again starting with the products we have two moles of ethanol in two moles carbon dioxide, two moles of ethanol, two moles of carbon dioxide. And this time we multiply each one of those Moeller values but their respective standard Moeller entropy values which we can look up in the appendix this time when we cancel out the units of moles we're left with finally units of jewels per Kelvin or total entropy of the products, we do the same for the reactant. Again, we just have that single mole of glucose multiplied by its standard Moeller entropy value. Give us the total entropy of the reactions. And now the total entropy of the products is this value and we subtract the total entropy of the reactant is to get the overall delta s of reaction be about 537 0.3 jewels per kelvin now again for Delta G. You could use either one of these and I'm going to choose to use this bottom equation and again. If you use this top one, you make it a slightly different answer, depending on the different information that is available. So Delta G. Eagles Delta H, which we calculated as negative 69 killer jewels before we subtract he the temperature, which is 298 Kelvin times Delta s which we just calculated. But we have to convert it into killer jewels per Kelvin by dividing by 1000 so that when we cancel off units of Kelvin, we're left with total Gibbs free energy change units of killer JAL's. And when we do all that mask, you get a final answer of about negative 229.1 killer jewels of free energy change. Now it In the second part of this problem we want to determine we want to know if the spontaneity of this reaction is dependent on temperature or not, we can examine this general equation again. Delta G equals still to h minus t Delta s. We assume that a Delta H in Delta s are independent of temperature and we calculated a negative value for Delta H the positive value for Delta s. We can look those in here Negative Delta H and a positive Delta s remember, for a reaction to be spontaneous, that means that it's value for Delta G. The change in Gibbs free energy is negative. And so we want to know if we changed the temperature. Will that well, any change of that temperature caused that value for Delta G to become positive. Remember that temperature is measured in absolute units of Kelvin and so the lowest that the temperature could possibly be zero. And if the temperature were zero where zero Kelvin in this product of the temperature in the entropy would become zero and we would be left with this constant negative value of the Delta H which would make this delta G value equal to that same quantity. And it would be a negative value, obviously. And so we can only increase the temperature above zero Kelvin, we cannot make we do not. There is no such thing as negative units for Kelvin. So if we increase that temperature, then this would become a more and more positive value when we want apply at positive temperature by the constant positive value for Delta s the term of tee times. Delta s becomes greater a greater positive value. We're subtracting a greater positive value from a constant negative value for Delta H. And so Delta H and Seau, Delta G can only therefore become more negative. And so because of that, we can see that whether we increase or decrease the temperature to its Teoh even its lowest possible value or if we increase it continuously, no matter what we do, the sign for Delta G will always be negative. And so the spontaneity is not dependent upon the temperature, meaning no matter how we change the temperature, the sign of Delta G will always be negative, and the reaction will therefore always be spontaneous.