Consider the following reaction:
CaCO3(s)→CaO(s)+CO2(g).
Estimate ΔG∘ for this reaction at each of the following temperatures. (Assume that ΔH∘ and ΔS∘ do not change too much within the given temperature range.)
A) 310K
B) 1070K
C) 1470K
And is A,B,C spontaneous or non spontaneous?
dHorxn = (n*dHoformation producs) - (n*dHoreactants)
dSo rxn = (n*dSo formation products( - (n*dSo formation reactants)
dGrxn = dHrxn - TdSrxn
dG<0 spontaneous
dG=0 equilibrium
dG>0 not spontaneous.
Im still confused..
is n 2?
and if so would it look like this for A
dHorxn=(2*2.84)-(2*-.227)??
n stands for number of mols. It is 1 for CaO, 1 for CaCO3, and 1 for CO2.
Look up dHo formation CO2, CaO, and CaCO3. Calculate dHo reaction.
Look up dSo for CO2, CaO, and CaCO3. Calculate dSo reaction.
Then dGorxn = dHorxn - TdSorxn
To estimate ΔG∘ for the given reaction at each temperature, we need to use the equation:
ΔG∘ = ΔH∘ - TΔS∘
Where:
ΔG∘ = standard Gibbs free energy change
ΔH∘ = standard enthalpy change
ΔS∘ = standard entropy change
T = temperature in Kelvin
We can obtain the values of ΔH∘ and ΔS∘ for the reaction from reference sources or experimental data. Let's assume the following values:
ΔH∘ = 178 kJ/mol
ΔS∘ = 160 J/(mol·K)
Now, let's calculate ΔG∘ at each temperature:
A) 310 K:
ΔG∘ = ΔH∘ - TΔS∘
= 178 kJ/mol - (310 K)(160 J/(mol·K))(10^-3 kJ/J)
= 178 kJ/mol - 49.6 kJ/mol
≈ 128.4 kJ/mol
B) 1070 K:
ΔG∘ = ΔH∘ - TΔS∘
= 178 kJ/mol - (1070 K)(160 J/(mol·K))(10^-3 kJ/J)
= 178 kJ/mol - 171.2 kJ/mol
≈ 6.8 kJ/mol
C) 1470 K:
ΔG∘ = ΔH∘ - TΔS∘
= 178 kJ/mol - (1470 K)(160 J/(mol·K))(10^-3 kJ/J)
= 178 kJ/mol - 235.2 kJ/mol
≈ -57.2 kJ/mol
Now, let's determine whether each reaction is spontaneous or non-spontaneous based on the sign of ΔG∘:
A) ΔG∘ = 128.4 kJ/mol
Since ΔG∘ is positive, the reaction at 310 K is non-spontaneous.
B) ΔG∘ = 6.8 kJ/mol
Since ΔG∘ is positive, the reaction at 1070 K is non-spontaneous.
C) ΔG∘ = -57.2 kJ/mol
Since ΔG∘ is negative, the reaction at 1470 K is spontaneous.