Oxygen reacts with the dipeptide glycylglycine [C4H8N2O3(s)] to form urea [CH4N2O], carbon dioxide and water:
3O2(g) + C4H8N2O3 (s) -> CH4N2O (s) + 3CO2(g) + 2H2O(l)
At T=298.15 K and 1.00atm, solid glycylglycine has the following thermodynamic properties:
DeltaGf= -491.5 kJ/mol DeltaHf= -746.0 kJ/mol S= 190.0 J/mol-K
i) Calculate DeltaGr at T=298K and T=310K. State any assumptions you make
To calculate the standard Gibbs free energy change (ΔGr) at a given temperature, you can use the equation:
ΔGr = ΔGr° + RT ln(Q)
where ΔGr° is the standard Gibbs free energy change at standard conditions, R is the gas constant (8.314 J/mol-K), T is the temperature in Kelvin, and Q is the reaction quotient.
Assumptions:
1. The reaction is at standard conditions (1 atm, 298.15 K).
2. The dipeptide glycylglycine is in its standard state.
Now, let's proceed to calculate ΔGr at T = 298.15 K and T = 310 K.
Step 1: Calculate ΔGr°
ΔGr° = ΣΔGf°(products) - ΣΔGf°(reactants)
From the balanced equation, the stoichiometric coefficients of the reactants and products are:
Reactants: 3O2(g) + C4H8N2O3 (s)
Products: CH4N2O (s) + 3CO2(g) + 2H2O(l)
Using the given standard Gibbs free energy of formation (ΔGf°) values:
ΔGf° (CH4N2O) = 0 kJ/mol (Urea)
ΔGf° (CO2) = -394.4 kJ/mol
ΔGf° (H2O) = -237.2 kJ/mol
ΔGf° (O2) = 0 kJ/mol (since it is an element)
ΔGf° (C4H8N2O3) = -491.5 kJ/mol (Glycylglycine)
ΔGr° = (1*0 kJ/mol + 3*(-394.4 kJ/mol) + 2*(-237.2 kJ/mol)) - (-491.5 kJ/mol)
ΔGr° = -1260.0 kJ/mol
Step 2: Calculate Q
In this case, since the reaction is at standard conditions, Q = 1.
Step 3: Calculate ΔGr at T = 298.15 K
ΔGr = ΔGr° + RT ln(Q)
= -1260.0 kJ/mol + (8.314 J/mol-K * 298.15 K * ln(1))
= -1260.0 kJ/mol
Therefore, ΔGr at T = 298.15 K is -1260.0 kJ/mol.
Step 4: Calculate ΔGr at T = 310 K
ΔGr = ΔGr° + RT ln(Q)
= -1260.0 kJ/mol + (8.314 J/mol-K * 310 K * ln(1))
= -1260.0 kJ/mol
Therefore, ΔGr at T = 310 K is also -1260.0 kJ/mol.
Note: Both ΔGr values are negative, indicating that the reaction is spontaneous at both temperatures.