Nitroglycerin is a powerful explosive, giving four different gases when detonated.2 C3H5(NO3)3 (l) → 3 N2 (g) + 1/2 O2 (g) + 6 CO2 (g) + 5 H2O (g)Given that the enthalpy of formation of nitroglycerin, ΔHf°, is −364 kJ/mol, calculate the energy (heat at constant pressure) released by this reaction.Δ= Σn􀁄rxnHfΔ􀁄H(prod.) − ΣmfΔ􀁄H(reac.)= (3 mol)(0) + (1/2 mol)(0) + (6 mol)(−393.5 kJ/mol) + (5 mol)(−241.8 kJ/mol) − (2 mol)(−364 kJ/mol) rxnΔ􀁄H= −2842 kJ rxnΔ􀁄Hb) What is delta H when 4.65 mol of products is formed?

Question

Nitroglycerin is a powerful explosive, giving four different gases when detonated.2 C3H5(NO3)3 (l) → 3 N2 (g) + 1/2 O2 (g) + 6 CO2 (g) + 5 H2O (g)Given that the enthalpy of formation of nitroglycerin, ΔHf°, is −364 kJ/mol, calculate the energy (heat at constant pressure) released by this reaction.Δ= Σn􀁄rxnHfΔ􀁄H(prod.) − ΣmfΔ􀁄H(reac.)= (3 mol)(0) + (1/2 mol)(0) + (6 mol)(−393.5 kJ/mol) + (5 mol)(−241.8 kJ/mol) − (2 mol)(−364 kJ/mol) rxnΔ􀁄H= −2842 kJ rxnΔ􀁄Hb) What is delta H when 4.65 mol of products is formed?

Answer 1

I'm not able to read all of the "funny" symbols you printed but if I've translated your question properly, I think the reaction produces 14.5 mols for 2842 kJ heat produced. You want 4.65 mols products so 2842 kJ/mol x (4.65/14.5) = ?

Answer 2

To find the energy released when 4.65 mol of products is formed, we need to calculate the change in enthalpy (∆H) for the given amount of products.

We can use the equation:
∆H = Σn∆Hf(prod.) - Σm∆Hf(reac.)

Here, ∆Hf(prod.) represents the enthalpy of formation of the products and ∆Hf(reac.) represents the enthalpy of formation of the reactants.

∆Hf(prod.) for N2 (g), O2 (g), CO2 (g), and H2O (g) are all zero, as these are considered to be in their standard states.

Similarly, we can find the values of ∆Hf(reac.) for the reactants C3H5(NO3)3 (l), which is -364 kJ/mol.

Using this information, we can substitute the values into the equation and calculate ∆H:
∆H = (3 mol)(0 kJ/mol) + (1/2 mol)(0 kJ/mol) + (6 mol)(-393.5 kJ/mol) + (5 mol)(-241.8 kJ/mol) - (4.65 mol)(-364 kJ/mol)

Calculating this expression gives us the value of ∆H for the given amount of products:
∆H = -2540.42 kJ

Therefore, the energy released (∆H) when 4.65 mol of products is formed is approximately -2540.42 kJ.

Answer 3

To calculate ΔH when 4.65 mol of products are formed, we can use the equation:

ΔH = ΔHrxn x (mol of products / mol of reaction)

Given that ΔHrxn = -2842 kJ, we can substitute the values into the equation:

ΔH = -2842 kJ x (4.65 mol / 2 mol)

ΔH = -2842 kJ x 2.325

ΔH ≈ -6613 kJ (rounded to the nearest whole number)

Therefore, the energy released when 4.65 mol of products are formed is approximately -6613 kJ.