Nitroglycerine, C3H5(NO3)3(l), is an explosive most often used in mine or quarry blasting. It is a powerful explosive because four gases (N2, O2, CO2, and steam) are formed when nitroglycerine is detonated. In addition, 6.26 kJ of heat is given off per gram of nitroglycerine detonated. (a) Write a balanced thermochemical equation for the reaction. (b) What is ΔH when 4.65 mol of products is formed?
molar mass C3H5N3O9 = 227
dH = 6.26 kJ/g x (227 g/mol) = 1421 kJ/mol
4C3H5(NO3)3 → 12CO2 + 6N2 + 10H2O + O2 dH = -1421 kJ/mol
29 mol products are formed for every 4 mol nitroglycerin detonated.
So 4.65 mol products must have come from 4.65 x (4/29) = 0.138 mol nitroglycerin.
1421 kJ/mol x 0.138 mol = ? kJ = dH for 4.65 mols products.
oops
Forgot to add the states. The thermochemical equation will be
4C3H5(NO3)3(l) → 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g) + 5684 kJ OR
4C3H5(NO3)3(l) → 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g) dH = -1421 kJ/mol OR
4C3H5(NO3)3(l) → 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g) dH = -5684 kJ
(a) To write a balanced thermochemical equation, we need to show the reactants and products of the reaction, as well as the energy change (ΔH) involved.
The balanced thermochemical equation for the detonation of nitroglycerine is:
4 C3H5(NO3)3(l) → 12 CO2(g) + 6 N2(g) + O2(g) + 7 H2O(g)
(b) To calculate ΔH when 4.65 mol of products is formed, we need to use the stoichiometric coefficients from the balanced equation and the given heat released per gram of nitroglycerine detonated.
From the balanced equation, we can see that for every 4 mol of nitroglycerine detonated, 12 mol of CO2, 6 mol of N2, 1 mol of O2, and 7 mol of H2O are formed.
Therefore, the mole ratio between nitroglycerine and products is:
4 mol nitroglycerine : 12 mol CO2 : 6 mol N2 : 1 mol O2 : 7 mol H2O
We can use this mole ratio to calculate the number of moles of products formed when 4.65 mol of nitroglycerine is detonated.
Number of moles of CO2 = (4.65 mol nitroglycerine) * (12 mol CO2 / 4 mol nitroglycerine) = 13.95 mol CO2
Number of moles of N2 = (4.65 mol nitroglycerine) * (6 mol N2 / 4 mol nitroglycerine) = 6.975 mol N2
Number of moles of O2 = (4.65 mol nitroglycerine) * (1 mol O2 / 4 mol nitroglycerine) = 1.1625 mol O2
Number of moles of H2O = (4.65 mol nitroglycerine) * (7 mol H2O / 4 mol nitroglycerine) = 8.1375 mol H2O
To calculate the total heat released, we can use the given heat released per gram of nitroglycerine detonated (6.26 kJ/g).
Total heat released = (4.65 mol nitroglycerine) * (227.09 g/mol) * (6.26 kJ/g) = 3383.64 kJ
To write a balanced thermochemical equation for the reaction, we need to understand the reactants and products involved. From the given information, we know that the reactant is nitroglycerine (C3H5(NO3)3(l)). When nitroglycerine is detonated, the products formed are N2, O2, CO2, and steam.
(a) Writing a balanced thermochemical equation:
The balanced equation for the reaction can be written as follows:
4C3H5(NO3)3(l) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)
It's important to note that the stoichiometric coefficients in the balanced equation are determined by the mole ratios obtained from the chemical formula of nitroglycerine.
(b) Finding ΔH for 4.65 mol of products:
To determine ΔH for the given number of moles of products, we'll first calculate the number of moles of reactants required using stoichiometry. From the balanced equation, we can see that the molar ratio of nitroglycerine to the products is 4:12:10:6:1. So for every 4 moles of nitroglycerine reacted, we have 12 moles of CO2, 10 moles of H2O, 6 moles of N2, and 1 mole of O2 produced.
For the given number of moles, we have:
4.65 mol CO2 = (4.65 mol C3H5(NO3)3) * (12 mol CO2 / 4 mol C3H5(NO3)3)
4.65 mol H2O = (4.65 mol C3H5(NO3)3) * (10 mol H2O / 4 mol C3H5(NO3)3)
4.65 mol N2 = (4.65 mol C3H5(NO3)3) * (6 mol N2 / 4 mol C3H5(NO3)3)
4.65 mol O2 = (4.65 mol C3H5(NO3)3) * (1 mol O2 / 4 mol C3H5(NO3)3)
Now, we multiply the number of moles by the enthalpy change per mole of the respective product. Given that the heat released is 6.26 kJ per gram of nitroglycerine detonated, it is necessary to convert grams of nitroglycerine to moles.
The molar mass of nitroglycerine (C3H5(NO3)3) = 227.09 g/mol
Converting grams to moles:
4.65 mol C3H5(NO3)3 = (4.65 mol C3H5(NO3)3) * (227.09 g/mol C3H5(NO3)3)
Finally, we calculate the enthalpy change:
ΔH = (4.65 mol CO2) * (6.26 kJ/mol) + (4.65 mol H2O) * (6.26 kJ/mol) + (4.65 mol N2) * (6.26 kJ/mol) + (4.65 mol O2) * (6.26 kJ/mol)
Performing the calculations will yield the value of ΔH for 4.65 mol of products.