Pentane (C5H12) is a component of gasoline that burns according
t...

Question

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.)

Accepted Answer

Hey there. What's up? All right, So here, I'm going to be calculating the delta H. Of this reaction that we have here. Now anytime we're dealing with reactions and we need to use the reaction to find our answers such as to Akiyama Tree or finding finding delta. Each of the reaction, we do need a balanced equation and it's always a good idea to actually check the reaction. So when you first look at this reaction, you'll see something is wrong, right? We have two carbons on the left side, but only one on the right side that stands out very easily, which means we actually do not have a balanced equation. So we do need to balance it. Let's go ahead and put to in front of the C. 02. Then we're going to need a three in front of 02 and then a three in front of each to oh, so now everything is balanced. Now, if we didn't check that, our answer would be completely wrong because we do need to know how many moles of each of the compound that were uh you know that we have in here because the delta transformations that are given to us are always individuals per one mole of it. And of course here we obviously do not have just one mole of each of those compounds. Okay, now let's get back to the calculation and let's see what equation we're going to be using for this. So to find delta age of reaction when we are dealing with delta H information of each of the compound. This is going to equal to excuse me? Delta Each information of the products minus the delta H information of the reactant. Okay, so we're just going to be plugging in numbers these numbers specifically and then of course we need to multiply them by the correct number of moles. Now you notice that actually oh to here is not given to us. Why is that? Because 02 is in its natural state? It's a di atomic molecules occurs like that naturally. So it's not going to have a delta H information associated with it. It's just zero and that's why we never include that. Alright, so let's go ahead and start. So, delta each. Actually I'm going to right over here so I don't run out over. And so delta H of this reaction, it's going to equal to. So first product is C. 02. We have two moles of that. So we write two moles of co two times. It's delta H information, which is negative 3 93. and killed joules per mole. Right? So we have a second product. So plus we have three moles of water times. It's delta H information As -285. killer joules per mole. Right, So that's products. Now we do minus the reactant again here for the reactant. We only have one And it's just one mole of it. Right, so one mole times um to delta H. A reaction for that is negative 2 77. kill jules per mold. Okay, just barely fit in here. Alright, so there we go. Let's go ahead and find the answer here. So for delta H reaction, When you input the numbers into the calculator to find the product side, you should get negative 1644 0. and this will be in killing jewels because notice that the molds here actually cancel out, right? And then minus the reactant, so minus a negative negative to 0.6 killer jewels. Okay, so that will give us an answer of negative 1366.8 kilo jewels. So you can write either kill or jewels or killed, joules per mole and that will just mean that it's per one mole of this reaction or per one mole of this reactant here that we have. Okay, because we have only one mole of that Per one mole of reaction. So writing either is correct. What you need to get here is obviously the correct value negative 1366.8 killed Jules. Alright folks, thank you so much for watching. Let us know if you have any questions

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.)

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