Calculating Standard Enthelpy

[TrueStar]

Homework Statement

4 C₃H₅N₃O₉ -> 12 CO₂+10 H₂O+6 N₂ -- H rxn = -5678kJ

Standard Heat of Formation of CO₂=-394KJ
Standard Heat of Formation of H₂O=-242KJ

Calculate the standard enthalpy of formation for nitroglycerin.

Homework Equations

Product minus reactant (I think)

The Attempt at a Solution

I thought you would swap this equation which gives 5678Kj. Then you have 12 moles of carbon dioxide and 10 moles of water (nitrogen is zero). Thus:

5678- (-4728 - 2420)

I think I'm way off though.

 

[nate808]

Your attempt at a solution is on the right track, but there are a few mistakes. First, the standard heat of formation for H2O is actually -286 kJ/mol, not -242 kJ/mol. Second, when calculating the standard enthalpy of formation, you need to use the coefficients of the balanced chemical equation. This means that you need to multiply the standard heat of formation for each product by its coefficient, and then subtract the sum of the standard heat of formation for each reactant multiplied by its coefficient. So the correct calculation would be:

ΔH°f = (12 mol CO2 * -394 kJ/mol) + (10 mol H2O * -286 kJ/mol) - (4 mol C3H5N3O9 * 0 kJ/mol)
= -4728 kJ + -2860 kJ + 0 kJ
= -7588 kJ/mol

So the standard enthalpy of formation for nitroglycerin is -7588 kJ/mol. This means that 7588 kJ of energy is released when 1 mol of nitroglycerin is formed from its elements in their standard states.

 

[Blueshift5]

Your attempt at a solution is close, but there are a few errors. First, the standard enthalpy of formation for CO2 and H2O should be -393.5 kJ/mol and -241.8 kJ/mol, respectively. In addition, the standard enthalpy of formation for N2 is 0 kJ/mol, not zero as you stated.

To calculate the standard enthalpy of formation for nitroglycerin, you need to use the equation:

ΔH°f(nitroglycerin) = ΣΔH°f(products) - ΣΔH°f(reactants)

where ΣΔH°f is the sum of the standard enthalpies of formation for each product or reactant.

In this case, the equation would look like:

ΔH°f(nitroglycerin) = (12 mol x -393.5 kJ/mol) + (10 mol x -241.8 kJ/mol) + (6 mol x 0 kJ/mol) - (-5678 kJ)

= -4722 kJ/mol

So the standard enthalpy of formation for nitroglycerin is -4722 kJ/mol.

Note: The values for the standard enthalpies of formation can vary slightly depending on the source, so it's important to use consistent values in your calculations.