The amino acid glycine, C2H5NO2, is one of the compounds used by the body to make proteins. The equation for its combustion is
4C2H4NO2(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 ΔHf° for the products of combustion, to calculate ΔHf° for glycine.
My answer:
973.49*4 = 3893.96kJ/mol
I don't think so.
dHrxn = (n*dHf products) - (n*dHf reactants)
4*-973.49 = (8*dHfCO2 + 10*dHf H2O) - (4*dHf glycine)
Solve for dHf glycine
To calculate ΔHf° for glycine, we need to use the equation provided and the values of ΔHf° for the products of combustion.
The equation for the combustion of glycine is:
4C2H4NO2(s) + 9O2(g) ---> 8CO2(g) + 10H2O(l) + 2N2(g)
Given that for each mole of glycine that burns, 973.49 kJ of heat is liberated, we can calculate the enthalpy change (ΔH) for the combustion reaction.
ΔH = 973.49 kJ/mol of glycine (1)
Now, let's examine the products of combustion:
- CO2: ΔHf° = -393.51 kJ/mol
- H2O: ΔHf° = -285.83 kJ/mol
- N2: ΔHf° = 0 kJ/mol (since it is an elemental form)
Using these values, we can calculate the enthalpy change (ΔH) for the formation of the products of combustion:
ΔH = (8 * ΔHf°CO2) + (10 * ΔHf°H2O) + (2 * ΔHf°N2)
Plugging in the values:
ΔH = (8 * -393.51 kJ/mol) + (10 * -285.83 kJ/mol) + (2 * 0 kJ/mol)
ΔH = -3148.08 kJ/mol - 2858.3 kJ/mol
ΔH = -6006.38 kJ/mol (2)
Since the combustion reaction is exothermic (heat is liberated), the enthalpy change (ΔH) for the reaction is negative.
Now, from the definition of heat of formation (ΔHf°), it is the enthalpy change for the formation of one mole of a compound from its elements in their standard states.
Using equation (1) and (2), we can write the following equation:
ΔHf°glycine = ΔH - (4 * ΔHf°CO2) - (5 * ΔHf°H2O)
Plugging in the values:
ΔHf°glycine = -6006.38 kJ/mol - (4 * -393.51 kJ/mol) - (5 * -285.83 kJ/mol)
ΔHf°glycine = -6006.38 kJ/mol + 1574.04 kJ/mol - 1429.15 kJ/mol
ΔHf°glycine = -5861.49 kJ/mol
Therefore, ΔHf° for glycine is approximately -5861.49 kJ/mol.
To calculate ΔHf° for glycine, we need to use the equation:
4C2H4NO2(s) + 9O2(g) ---> 8CO2(g) + 10H2O(l) + 2N2(g)
We know that for each mole of glycine burned, 973.49 kJ of heat is liberated. So, we can multiply the heat evolved by the number of moles of glycine:
973.49 kJ/mol × 4 mol = 3893.96 kJ/mol
Therefore, ΔHf° for glycine is 3893.96 kJ/mol.