What is ΔH for the explosion of nitroglycerin? (LO 9.14)
2 C3H5(N...

Question

What is ΔH for the explosion of nitroglycerin? (LO 9.14)
2 C3H5(NO3)3(l) → 3 N2(g) + 1/2 O2(g) + 6 CO2(g) + 5 H2O(g)
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(a) −315.0 kJ
(b) −4517 kJ
(c) −3425 kJ
(d) −3062 kJ

Accepted Answer

hi everyone for this problem we're told that the combustion of fennel under standard conditions is shown below were also given the standard entropy change for this reaction. We need to calculate the standard heat of formation for fennel. So in order for us to solve this problem, we're first going to need to look up the standard heats of formation for everything that's in our problem. So let's go ahead and do that first. The standard heat of formation for liquid water is negative 285.8 kg jewels her more. The standard heat of formation of carbon dioxide gas is negative 393.5 Killer jewels Permal. And our standard heat of formation for oxygen gas. And something to remember is that anything in its elemental state is going to have a standard heat of formation of zero. So this is going to be zero killer joules per mole. And so here we need to solve for the standard heat of formation for fennel. So our reaction or equation that we're going to use to solve this is the standard entropy change of our reaction is going to equal the sum of our products minus the sum of our reactant. And what we're going to do here is plug in our values for our standard heats of formation and multiply it by the number of moles we have. And so for our standard entropy change for the reaction they gave it to us so we can go ahead and plug that in. They tell us that it's negative killer jules Permal. So that's our left side. Now that's going to be equal to the sum of our products minus the sum of our reactant. So for our products we have three moles of liquid water. So our three moles of Liquid water multiplied by its standard heat of formation, which we said is negative 0.8 killer jewels per mole. Plus We have six moles of carbon dioxide gas, multiplied by its standard heat of formation, negative 393.5 killer jewels per mole. So that's the sum of our products and we're going to minus the sum of our reactant. And so for our reactant we have six moles, Sorry, we have seven moles of oxygen gas and we said that anything in its elemental state is going to be zero. So that's going to be multiplied by zero killer jewels per mole. And then we have plus one mole of fennel, but we don't have its standard heat of formation. That's what we're solving for. So we can say that that's multiplied by X. We're multiplying it by X. Because we need to solve for X. Which is going to tell us what is that standard heat of formation. So let's go ahead and simplify this out. And when we do, we get negative 3050 for killer jewels per mole is equal to -3, 0.4 kg joules per mole minus one. X. Okay, so here we know that this seven times zero is going to cancel. And so when we simplify it out, this is what we get. So now we're going to solve for X. So we can do that by adding, we're going to add And 18.4 to both sides. And when we do that and simplify, we're going to get On the left side 164.4 killer jewels. Permal is going to equal negative one. X. So now we can go ahead and divide a negative one And when we do that we're going to get X is equal to negative 164 0. killer jewels per mole. And this is going to be equal to our standard Heat of formation for Fennel, which is C6 H H. This is our final answer and that is the end of this problem. I hope this was helpful

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