Glycine is the amino acid used to make proteins
a)for each mole of glycine that burns,973.49k/mol of which is libarated eqution and the values of delta Hf^o for the products of combustion to calculate the enthalpy of formation for glycine.Represent your working on the energy diagram
To calculate the enthalpy of formation for glycine (ΔHf°), we need the balanced chemical equation for its combustion and the values of ΔHf° for the products involved. The balanced combustion equation for glycine is:
C2H5NO2 + 5O2 → 3CO2 + 3H2O + N2
First, let's write down the given information:
- The molar ratio between glycine burnt and energy liberated is 1 mole glycine: 973.49 kJ/mol.
- We need the ΔHf° values for CO2, H2O, and N2.
Now, let's calculate the enthalpy of formation for glycine step by step:
1. Write down the equation for the combustion of glycine, including the ΔHf° values:
C2H5NO2 + 5O2 → 3CO2 + 3H2O + N2
ΔHf° = ? kJ/mol
2. Use the balanced equation to relate the energy liberated by 1 mole of glycine burnt to the enthalpy of formation:
ΔHf°(glycine) = [ΔHf°(CO2) * 3] + [ΔHf°(H2O) * 3] + [ΔHf°(N2)] - 5[ΔHf°(O2)] - 973.49 kJ/mol
3. Substitute the known values of ΔHf° for CO2, H2O, and N2. These values can be found in reference tables or through thermodynamic databases.
For example, assuming:
ΔHf°(CO2) = -393.5 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol
ΔHf°(N2) = 0 kJ/mol
ΔHf°(O2) = 0 kJ/mol
The equation becomes:
ΔHf°(glycine) = [-393.5 kJ/mol * 3] + [-285.8 kJ/mol * 3] + [0 kJ/mol] - 5[0 kJ/mol] - 973.49 kJ/mol
4. Perform the calculations and simplify to find the enthalpy of formation for glycine.
ΔHf°(glycine) = -1180.5 kJ/mol + (-857.4 kJ/mol) - 973.49 kJ/mol
ΔHf°(glycine) = -3011.39 kJ/mol
The negative sign indicates that the formation of glycine is exothermic, releasing energy during the process.
Finally, to represent this working on an energy diagram, you can draw a diagram with the reactants on the left side, the products on the right side, and label the energy change (ΔHf°) associated with each step of the reaction.