The amino acid glycine, C2H5NO2, is one of the compounds used by the body to make proteins. The
equation for its combustion is
4C2H5NO2(s) + 9O2(g) -------------> 8CO2(g) + 10H2O(l) + 2N2(g)
For each mole of glycine that burns, 973.49 kJ of heat is liberated. Use this information plus values of Hof for the products of combustion to calculate Hof
for glycine.
To calculate the standard enthalpy of formation (∆Hof) for glycine (C2H5NO2), you need to use the equation:
∆Hof(glycine) = ∑∆Hof(Products) - ∑∆Hof(Reactants)
First, you need to find the ∆Hof values for the products and reactants using data from standard reference tables or a reliable source. The ∆Hof values represent the enthalpy change when 1 mole of a compound is formed from its constituent elements under standard conditions (25°C and 1 atm pressure).
Using the given equation for combustion:
4C2H5NO2(s) + 9O2(g) -> 8CO2(g) + 10H2O(l) + 2N2(g)
The reactants are 4 moles of glycine and 9 moles of oxygen gas. The products are 8 moles of carbon dioxide, 10 moles of water, and 2 moles of nitrogen gas.
Next, look up the ∆Hof values for each of the products and reactants. For example:
∆Hof(CO2) = -393.51 kJ/mol (standard enthalpy of formation of carbon dioxide)
∆Hof(H2O) = -285.83 kJ/mol (standard enthalpy of formation of water)
∆Hof(N2) = 0 kJ/mol (standard enthalpy of formation of nitrogen gas)
Note: The values given for ∆Hof are negative because the combustion products are more stable than the reactants.
Now, substitute these values into the equation:
∆Hof(glycine) = [8 * ∆Hof(CO2)] + [10 * ∆Hof(H2O)] + [2 * ∆Hof(N2)] - [4 * ∆Hof(glycine)] - [9 * ∆Hof(O2)]
∆Hof(glycine) = [8 * (-393.51 kJ/mol)] + [10 * (-285.83 kJ/mol)] + [2 * 0 kJ/mol] - [4 * ∆Hof(glycine)] - [9 * 0 kJ/mol]
∆Hof(glycine) = -3148.08 kJ/mol - 2858.3 kJ/mol - 0 kJ/mol - [4 * ∆Hof(glycine)] - 0 kJ/mol
Now, rearrange the equation to solve for ∆Hof(glycine):
[4 * ∆Hof(glycine)] = -3148.08 kJ/mol - 2858.3 kJ/mol
[4 * ∆Hof(glycine)] = -6006.38 kJ/mol
∆Hof(glycine) = (-6006.38 kJ/mol) / 4
∆Hof(glycine) = -1501.6 kJ/mol
Therefore, the ∆Hof for glycine is approximately -1501.6 kJ/mol.
dHrxn = 973.49 kJ/mol x 4 = ?
dHrxn = (n*dHof products) - (n*dHof reactants)
You know dHrxn dHo of all but glycine. Substitute and solve for dHof glycine.