The explosive nitroglycerin (C3H5N3O9) decomposes rapidly
upon ig...

Question

The explosive nitroglycerin (C3H5N3O9) decomposes rapidly
upon ignition or sudden impact according to the balanced
equation:
4 C3H5N3O9(l )¡12 CO2( g) + 10 H2O( g) + 6 N2( g) + O2( g)
ΔH °rxn = -5678 kJ
Calculate the standard enthalpy of formation (ΔH °f ) for
nitroglycerin.

Accepted Answer

Hi everyone for this problem, we're told that tri nitro towline or more commonly known as TNT is used as a high explosive for military and industrial applications. The reaction for its explosion is shown below and they also give us the standard entropy change of the reaction. We need to calculate the standard heat of formation for T. N. T. So let's get started. In order to solve this problem, we need to know that the standard entropy change of our reaction is going to equal the sum of our products minus the sum of our reactant. And I will explain what that means. Okay, so this standard entropy change for our reaction is given, they tell us that it is negative 20306 killer jewels. And this is going to equal the standard heat of formation of our products minus the standard heat of formation of our reactant. And so we need to pay attention to what is in our equation. Anything in its elemental state is going to equal zero. So here we have three things in their elemental state. We have and two gas, H two gas and we have a solid carbon here. So all of those are in their standard states. And so the delta age of formation or a standard entropy change of formation for anything in its elemental state is going to equal zero. So we know that we're not going to pay attention to those three products. So we do have one product here and it's 12 moles of C. O. Gas. And we need to look up the standard entropy or the heat of formation of C. O. Gas. And when we do it is equal to of our C. O gas. This is equal to negative .5 Killer Jewels Per Mole. So we can go ahead and plug that into our equation, paying attention to how many moles of it we have. So we have 12 moles of c. o. gas and its standard heat of formation is negative 110.5 kila jewels per mole. So that's our products because we're doing the some of our products minus the sum of our reactant and so we only had one product, so this is going to be minus our reactant. And here we have one reactant which is our TNT, what we're trying to find out and we know we have two moles of it, but we don't know what the standard heat of formation for it is, That's what we're solving for. So we can go ahead and put that as X. So essentially we're solving for X here. So When we go ahead and simplify this out, we'll get negative 20, 306 Killer Jewels is equal to negative 1326 killer jewels per mole minus two X. And remember we are solving for this X. Because that will give us our our standard heat of formation for TNT. Okay, so let's go ahead and add 1326 kg joules per mole to both sides. And when we do that, we will get on the left side -1000 or 18, 980 killer jewels per mole is going to equal negative two X. We will divide by negative two and we will get a final answer of X is equal to 9490 killer jewels per more. And this is going to be our standard heat of formation for our TNT. That's the end of this problem. I hope this was helpful.

The explosive nitroglycerin (C3H5N3O9) decomposes rapidly upon ignition or sudden impact according to the balanced equation: 4 C3H5N3O9(l )¡12 CO2( g) + 10 H2O( g) + 6 N2( g) + O2( g) ΔH °rxn = -5678 kJ Calculate the standard enthalpy of formation (ΔH °f ) for nitroglycerin.

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