A2.27g sample of the explosive, nitroglycerin, C3H5(NO3)3, is allowed to explode in a bomb calorimeter at standard conditions. The amount of heat evolved is 15.1 kJ. This explosion of nitroglycerin occurs according to the equation:
2C3H5(NO3)3 (l)⟶6CO2 (g) + 5H2O (l) + 3N2 (g) +1/2O2 (g)
calculate ΔHf for nitroglycerin. (note that the "f" refers to formation)
dHrxn = (n*dHf products) - (n*dHf reactants)
Set dHrxn to -15.1 kJ and solve for dHf for nitroglycerin.
To calculate the enthalpy of formation (ΔHf) for nitroglycerin (C3H5(NO3)3), we need to use the given equation for the explosion of nitroglycerin and the enthalpy values of formation for the other substances involved.
The enthalpy change for a reaction can be calculated using the following equation:
ΔH = ΣnΔHf(products) - ΣmΔHf(reactants)
Where ΔH is the enthalpy change, ΣnΔHf(products) is the sum of the products' enthalpies of formation multiplied by their stoichiometric coefficients, and ΣmΔHf(reactants) is the sum of the reactants' enthalpies of formation multiplied by their stoichiometric coefficients.
In this case, we need to find ΔHf for nitroglycerin, so we rearrange the equation as follows:
ΔH = ΣnΔHf(products) - ΣmΔHf(reactants) ==> ΔH + ΣmΔHf(reactants) = ΣnΔHf(products)
Now let's calculate ΔH using the given values of the enthalpy change and the enthalpies of formation of the other substances involved:
Given:
Amount of heat evolved (ΔH) = 15.1 kJ
Enthalpy of formation for products:
ΔHf(CO2) = -393.5 kJ/mol
ΔHf(H2O) = -285.8 kJ/mol
ΔHf(N2) = 0 kJ/mol
ΔHf(O2) = 0 kJ/mol
The stoichiometric coefficients for the products are:
CO2: 6
H2O: 5
N2: 3
O2: 1/2
The stoichiometric coefficients for nitroglycerin in the equation are:
C3H5(NO3)3: 2
Now, substitute the given values into the equation and solve for ΔHf for nitroglycerin:
ΔH + (2ΔHf(C3H5(NO3)3)) = (6ΔHf(CO2)) + (5ΔHf(H2O)) + (3ΔHf(N2)) + (1/2ΔHf(O2))
15.1 kJ + (2ΔHf(C3H5(NO3)3)) = (6 * -393.5 kJ/mol) + (5 * -285.8 kJ/mol) + (3 * 0 kJ/mol) + (1/2 * 0 kJ/mol)
15.1 kJ + (2ΔHf(C3H5(NO3)3)) = -2361 kJ/mol - 1429 kJ/mol
15.1 kJ + (2ΔHf(C3H5(NO3)3)) = -3790 kJ/mol
Now, solve for ΔHf(C3H5(NO3)3):
2ΔHf(C3H5(NO3)3) = -3790 kJ/mol - 15.1 kJ
2ΔHf(C3H5(NO3)3) = -3805.1 kJ/mol
ΔHf(C3H5(NO3)3) = -1902.5 kJ/mol
Therefore, the enthalpy of formation for nitroglycerin (C3H5(NO3)3) is approximately -1902.5 kJ/mol.