2CH3CH2CH2OH(l)+9O2(g)→6CO2(g)+8H2O(g)
I have to find the delta G
delta G of CH3CH2CH2OH(l)=−360.5 KJ/Mol
delta G of O2(g)=0 KJ/Mol
delta G of CO2(g)=−394.4 KJ/Mol
delta G of H20(g)=−228.6 KJ/Mol
I keep getting -3426.2 KJ/Mol but its not correct.
Hess Law:
sum heats products-sum reactants
8*-228.6 + 6*-394.4-2*360.5=
which does not match your answer.
Oh it sums up the heat? I thought it was delta G(products)-Delta G(reactants) wouldn't that then be
8*-228.6 + 6*-394.4-2*(-360.5)= giving you -3426.2 KJ/Mol. Could you explain why you made the 360.5 positive?
Thanks.
Because I made a typo error on the sign on the 360, and you have the correct procedure. It is -3426.2Kj
I am doing this assignment for mastering chemistry and It just keeps telling me that it is wrong, I have no idea what the issue is.
To find the standard Gibbs free energy change (ΔG°) for a reaction, you need to use the thermodynamic data for each compound involved in the reaction. The ΔG° for the reaction can be calculated by subtracting the sum of the ΔG° values of the reactants from the sum of the ΔG° values of the products.
Given the following values:
ΔG° of CH3CH2CH2OH(l) = -360.5 kJ/mol
ΔG° of O2(g) = 0 kJ/mol
ΔG° of CO2(g) = -394.4 kJ/mol
ΔG° of H2O(g) = -228.6 kJ/mol
The balanced reaction equation is:
2CH3CH2CH2OH(l) + 9O2(g) → 6CO2(g) + 8H2O(g)
First, calculate the sum of the ΔG° values of the reactants:
2 × ΔG° of CH3CH2CH2OH(l) = 2 × (-360.5 kJ/mol) = -721 kJ/mol
9 × ΔG° of O2(g) = 9 × (0 kJ/mol) = 0 kJ/mol
Next, calculate the sum of the ΔG° values of the products:
6 × ΔG° of CO2(g) = 6 × (-394.4 kJ/mol) = -2366.4 kJ/mol
8 × ΔG° of H2O(g) = 8 × (-228.6 kJ/mol) = -1828.8 kJ/mol
Then, subtract the sum of the reactant values from the sum of the product values:
ΔG° = (-721 kJ/mol + 0 kJ/mol) - (-2366.4 kJ/mol + -1828.8 kJ/mol)
= 721 kJ/mol + 2366.4 kJ/mol + 1828.8 kJ/mol
= 4916.2 kJ/mol
Therefore, the correct value for the standard Gibbs free energy change (ΔG°) of the given reaction is 4916.2 kJ/mol.