Calculate Delta G at 298 K for the following reactions.
Ca(s) + CO2(g) + 1/2O2(g) -> CaCO3(s)
2KClO3(g) -> 2KCl(s) + 3O2(g)
CH4(g) + 3Cl2(g) -> CHCl3(g) + 3HCl(g)
Delta G^0f CHCl3 is - 70.4 kJ/mol
The following reactions are important ones in catalytic converters in automobiles. Calculate Delta G for each at 298 K.
2CO2(g) + 2NO(g) -> N2(g) + 2CO2(g)
2NH3(g) + 2O2(g) -> N2O(g) + 3H2O(g)
For the first reaction:
Delta G = - 572.4 kJ/mol
For the second reaction:
Delta G = - 990.2 kJ/mol
To calculate the standard Gibbs free energy change, ΔG°, at 298 K for a given reaction, we need to use the equation:
ΔG° = ΣnΔG°f(products) - ΣnΔG°f(reactants)
where ΔG°f is the standard Gibbs free energy of formation, n is the stoichiometric coefficient, and the summations are taken over all the reactants and products.
Let's calculate ΔG° for each of the given reactions:
1. Ca(s) + CO2(g) + 1/2O2(g) -> CaCO3(s)
To calculate ΔG°, we need the standard Gibbs free energy of formation for each compound involved in the reaction. We can look up these values in a table of standard values. Let's assume the values are as follows:
ΔG°f(CaCO3) = -1206 kJ/mol
ΔG°f(Ca) = 0 kJ/mol
ΔG°f(CO2) = -394 kJ/mol
ΔG°f(O2) = 0 kJ/mol
Using these values, we can calculate:
ΔG° = (1 * ΔG°f(CaCO3)) - [(1 * ΔG°f(Ca)) + (1 * ΔG°f(CO2)) + (1/2 * ΔG°f(O2))]
ΔG° = (1 * -1206) - [(1 * 0) + (1 * -394) + (1/2 * 0)]
ΔG° = -1206 + 0 + (-394/2)
ΔG° = -1206 - 197
ΔG° = -1403 kJ/mol
Therefore, ΔG° for the first reaction is -1403 kJ/mol at 298 K.
2. 2KClO3(g) -> 2KCl(s) + 3O2(g)
Let's assume the following values:
ΔG°f(KClO3) = -394 kJ/mol
ΔG°f(KCl) = -436 kJ/mol
ΔG°f(O2) = 0 kJ/mol
Using these values, we can calculate:
ΔG° = [(2 * ΔG°f(KCl)) + (3 * ΔG°f(O2))] - (2 * ΔG°f(KClO3))
ΔG° = [(2 * -436) + (3 * 0)] - (2 * -394)
ΔG° = (-872 + 0) - (-788)
ΔG° = -872 + 788
ΔG° = -84 kJ/mol
Therefore, ΔG° for the second reaction is -84 kJ/mol at 298 K.
3. CH4(g) + 3Cl2(g) -> CHCl3(g) + 3HCl(g)
The given reaction does not provide the ΔG°f values for CH4 and Cl2, so we cannot directly calculate ΔG°. However, if we have the ΔG°f value for CHCl3, we can calculate ΔG° using Hess's Law.
Let's assume the following values:
ΔG°f(CHCl3) = -70.4 kJ/mol
ΔG°f(HCl) = -95.3 kJ/mol
Since ΔG°f(CH4) and ΔG°f(Cl2) are not given, we need to consider the reaction by parts:
a) CH4(g) -> C(s) + 2H2(g)
b) Cl2(g) -> 2 Cl(g)
Using the ΔG°f values for C, H2, and Cl(g), we can calculate:
ΔG°a = [(1 * ΔG°f(C)) + (2 * ΔG°f(H2))] - (1 * ΔG°f(CH4))
ΔG°a = [(1 * 0) + (2 * 0)] - (1 * ΔG°f(CH4))
ΔG°a = - ΔG°f(CH4)
Similarly, for reaction b:
ΔG°b = [(2 * ΔG°f(Cl))] - [(1 * ΔG°f(Cl2))]
ΔG°b = [(2 * 0)] - [(1 * 0)]
ΔG°b = 0
Now, using ΔG°a, ΔG°b, and ΔG°f(CHCl3), we can calculate ΔG° for the full reaction:
ΔG° = [(1 * ΔG°f(CHCl3)) + (3 * ΔG°f(HCl))] - ΔG°a - 3ΔG°b
ΔG° = [(1 * -70.4) + (3 * -95.3)] - (- ΔG°f(CH4)) - 3(0)
ΔG° = (-70.4 - 285.9) - (- ΔG°f(CH4))
ΔG° = -70.4 - 285.9 + ΔG°f(CH4)
Therefore, ΔG° for the third reaction cannot be determined without knowing the ΔG°f value of CH4.
4. 2CO2(g) + 2NO(g) -> N2(g) + 2CO2(g)
Let's assume the following values:
ΔG°f(CO2) = -394 kJ/mol
ΔG°f(NO) = 86 kJ/mol
ΔG°f(N2) = 0 kJ/mol
Using these values, we can calculate:
ΔG° = [(1 * ΔG°f(N2))] - [(2 * ΔG°f(CO2)) + (2 * ΔG°f(NO))]
ΔG° = (1 * 0) - [(2 * -394) + (2 * 86)]
ΔG° = 0 - [-788 + 172]
ΔG° = 0 + 616
ΔG° = 616 kJ/mol
Therefore, ΔG° for the fourth reaction is 616 kJ/mol at 298 K.
5. 2NH3(g) + 2O2(g) -> N2O(g) + 3H2O(g)
Let's assume the following values:
ΔG°f(NH3) = -45.9 kJ/mol
ΔG°f(O2) = 0 kJ/mol
ΔG°f(N2O) = 82.05 kJ/mol
ΔG°f(H2O) = -228.6 kJ/mol
Using these values, we can calculate:
ΔG° = [(1 * ΔG°f(N2O)) + (3 * ΔG°f(H2O))] - [(2 * ΔG°f(NH3)) + (2 * ΔG°f(O2))]
ΔG° = [(1 * 82.05) + (3 * -228.6)] - [(2 * -45.9) + (2 * 0)]
ΔG° = (82.05 - 685.8) - (-91.8 + 0)
ΔG° = -603.75 + 91.8
ΔG° = -511.95 kJ/mol
Therefore, ΔG° for the fifth reaction is -511.95 kJ/mol at 298 K.
To calculate Delta G at 298 K for the given reactions, we need to use the standard Gibbs free energy change values (Delta G^0) for each reaction. These values can be obtained from thermodynamic tables.
1. Ca(s) + CO2(g) + 1/2O2(g) -> CaCO3(s)
The standard Gibbs free energy change for this reaction is not given. However, we can calculate it using the standard Gibbs free energy change values for formation of each compound involved in the reaction. The equation we can use is:
Delta G^0 = (Sum of Delta G^0f of products) - (Sum of Delta G^0f of reactants)
The standard Gibbs free energy of formation (Delta G^0f) at 298 K for each compound involved in the reaction is:
Delta G^0f for CaCO3(s) = -1128.3 kJ/mol
Delta G^0f for Ca(s) = 0 kJ/mol
Delta G^0f for CO2(g) = -394.4 kJ/mol
Delta G^0f for O2(g) = 0 kJ/mol
Therefore, substituting the values:
Delta G^0 = (-1128.3) - 0 - 394.4 - 0
= -1522.7 kJ/mol
2. 2KClO3(g) -> 2KCl(s) + 3O2(g)
The standard Gibbs free energy change for this reaction can be determined directly from the standard Gibbs free energy change values for each reactant and product. The equation we can use is:
Delta G^0 = (Sum of Delta G^0 of products) - (Sum of Delta G^0 of reactants)
The standard Gibbs free energy of formation (Delta G^0f) at 298 K for each compound involved in the reaction is:
Delta G^0f for KClO3(g) = -225.5 kJ/mol
Delta G^0f for KCl(s) = -415.1 kJ/mol
Delta G^0f for O2(g) = 0 kJ/mol
Therefore, substituting the values:
Delta G^0 = (0 + 0) - (-225.5 - 415.1)
= 640.6 kJ/mol
3. CH4(g) + 3Cl2(g) -> CHCl3(g) + 3HCl(g)
The standard Gibbs free energy change for this reaction can be determined directly from the standard Gibbs free energy change values for each reactant and product. The equation we can use is:
Delta G^0 = (Sum of Delta G^0 of products) - (Sum of Delta G^0 of reactants)
The standard Gibbs free energy of formation (Delta G^0f) at 298 K for each compound involved in the reaction is:
Delta G^0f for CH4(g) = -50.8 kJ/mol
Delta G^0f for CHCl3(g) = -70.4 kJ/mol
Delta G^0f for Cl2(g) = 0 kJ/mol
Delta G^0f for HCl(g) = -92.3 kJ/mol
Therefore, substituting the values:
Delta G^0 = (-70.4 + (3 * -92.3)) - ((-50.8) + (3 * 0))
= -111.9 kJ/mol
For the reactions related to catalytic converters in automobiles:
1. 2CO2(g) + 2NO(g) -> N2(g) + 2CO2(g)
Since the reaction involves the same compounds on both sides, the standard Gibbs free energy change will be zero (Delta G^0 = 0).
2. 2NH3(g) + 2O2(g) -> N2O(g) + 3H2O(g)
The standard Gibbs free energy change for this reaction can be determined directly from the standard Gibbs free energy change values for each reactant and product. The equation we can use is:
Delta G^0 = (Sum of Delta G^0 of products) - (Sum of Delta G^0 of reactants)
The standard Gibbs free energy of formation (Delta G^0f) at 298 K for each compound involved in the reaction is:
Delta G^0f for NH3(g) = -16.5 kJ/mol
Delta G^0f for N2O(g) = 82.05 kJ/mol
Delta G^0f for O2(g) = 0 kJ/mol
Delta G^0f for H2O(g) = -241.82 kJ/mol
Therefore, substituting the values:
Delta G^0 = (82.05 + (3 * -241.82)) - ((2 * -16.5) + (2 * 0))
= -1459.43 kJ/mol