Pentane (C5H12) is a component of gasoline that burns according to the following balanced equation: C5H12(l ) + 8 O2( g)¡5 CO2( g) + 6 H2O( g) Calculate ΔH °rxn for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of liquid pentane is -146.8 kJ>mol.)

Video transcript

All right. Hi, everyone. So this question says that hept at C seven H 16 burns according to the following balanced equation. One more of hep react with 11 moles of oxygen gas. They yield seven moles of carbon dioxide and eight moles of water vapor. Calculate delta H standard for this reaction using standard entropies of formation. The standard entropy of formation of liquid pentane is equal to negative 223.0 keto joules per mole. And here we have four different answer choices proposing different values for delta H standard of the reaction in units of kilo Juel. Now recall that for this question, we're going to have to recall the standard entropies of formation for all reactants and all products. And this is because the standard entropy of the reaction can be calculated by taking the difference between the standard entropy formation of all products and the standard entropy of formation of all the reactants. So to solve this question, we're going to add the standard entropies of formation of the products and the reactants and then find the difference between both of those terms. Now this requires standard entropies of formation that can be found in a standard table of your textbook. So here in addition to the standard entropy of formation or tain, the standard entropy of formation for carbon dioxide gas is equal to negative 393.5 kilo joules per mole. The standard enthalpy a formation for water vapor is equal to negative 241.8 kilo jewels per M. And finally, the standard entropy formation for oxygen gas is actually 0 kg joules per mole as is the case for all elements in their standard state. So at this point, we can go ahead and use these entropy values to find delta H standard for the reaction. So first I am going to go ahead and find the sum of the standard entropies of formation of all the products. So referencing our balanced chemical equation, our products are carbon dioxide gas and water vapor. So I'm going to add together their standard entropies of formation but also mort of pie by the number of moles of each in the balanced chemical equation. So starting off with carbon dioxide gas, I would multiply the number of moles of carbon dioxide that's seven buy the standard enthropy formation which is negative 393.5 kilo jules Perma. So then I will add this to the number of moles of water vapor that's eight multiplied by its standard entropy of formation. So for water that's negative 241.8 kilo joules per month. So after combining all of these terms, the standard entropy of formation for all products is equal to negative 4688.9 kg jewels. So now we're going to do the same thing. But for all reactants, delta H standard formation for reactants is equal to the sum of the standard ops of formation or the reactants. In this case, oxygen gas and heptane multiplied buy their motor coefficients. But because oxygen gas is in its standard state, its standard entropy of formation is going to be zero. So it is not going to contribute to the total right, since they would simply cancel out a zero. So in this case for the delta H standard formation of the reactants, I'm going to use that of HEPT multiplied by its molar coefficient. So according to the balanced chemical equation, I have only one mole of hep. So I multiply that by its standard entropy of formation, which is negative 200 23.0 kilo joules per m. And so this simply equals negative 223.0 kg joules. And so our last step is two subtract the total entropies of the products by the total entropies of the reactants. So delta H standard of the reaction is equal to negative 4688.9 kg joules subtracted by negative 223.0 kg joules and this equals negative 4465.9 kg jules, which corresponds to our final answer of option D in the multiple choice and there you have it. So if you watch this video all the way to the end, thank you so very much. I appreciate it. And I hope you found this helpful.