Use the data in the table to calculate the Delta G0 of the following reaction at 298k:
Al2O3 ---> Delta G fusion -1582.3
HCl----> -131.2
AlCl3 -----628.8
H2O --- -237.1
The equation is Al2O3 + 6HCl ---> 2 AlCl3 + 3H2O
products - reactants?
[2(-628.8)+3(-237.1) - (-1582.3)+6(-131.2) =-1178.8 KJ
But the answer is supposed to be +400.6 KJ/mol
If you do the math right you get 400.6 kJ/mol. At least that's what I obtained.
400
Well, it seems like there might have been a mix-up with the calculations. But let me lighten up the mood a bit with a joke before we continue:
Why don't scientists trust atoms?
Because they make up everything!
Now, let's get back to business. To calculate the ΔG° of the reaction, we need to consider the ΔG° values of the products and subtract the sum of the ΔG° values of the reactants.
ΔG° = [2*(-628.8) + 3*(-237.1)] - [(-1582.3) + 6*(-131.2)]
After simplifying the equation, we find:
ΔG° = 400.6 kJ/mol
So, despite the initial confusion, the correct answer is indeed +400.6 kJ/mol. Just goes to show that even in chemistry, things can sometimes have unexpected outcomes!
To calculate the Gibbs free energy change (ΔG°) of a reaction, you need to use the equation:
ΔG° = ΣΔG°(products) - ΣΔG°(reactants)
Given the following values:
ΔG° fusion of Al2O3 = -1582.3 kJ/mol
ΔG° of HCl = -131.2 kJ/mol
ΔG° of AlCl3 = 628.8 kJ/mol
ΔG° of H2O = -237.1 kJ/mol
Using the equation: Al2O3 + 6HCl ---> 2AlCl3 + 3H2O
Let's calculate the ΔG°:
ΣΔG°(products) = 2(-628.8) + 3(-237.1) = -1257.6 - 711.3 = -1968.9 kJ/mol
ΣΔG°(reactants) = -1582.3 + 6(-131.2) = -1582.3 - 787.2 = -2369.5 kJ/mol
ΔG° = -1968.9 - (-2369.5) = -1968.9 + 2369.5 = 400.6 kJ/mol
Therefore, the ΔG° for the reaction at 298K is +400.6 kJ/mol.
To calculate the ΔG0 of the reaction, you need to use the formula:
ΔG0 = ΣΔG0(products) - ΣΔG0(reactants)
Given the following information in the table:
ΔG0 fusion of Al2O3 = -1582.3 kJ/mol
ΔG0 of HCl = -131.2 kJ/mol
ΔG0 of AlCl3 = 628.8 kJ/mol
ΔG0 of H2O = -237.1 kJ/mol
The balanced equation for the reaction is:
Al2O3 + 6HCl ---> 2AlCl3 + 3H2O
Now, substitute the values into the equation:
ΔG0 = [2(ΔG0 of AlCl3) + 3(ΔG0 of H2O)] - [ΔG0 of Al2O3 + 6(ΔG0 of HCl)]
ΔG0 = [2(628.8) + 3(-237.1)] - [-1582.3 + 6(-131.2)]
ΔG0 = [1257.6 - 711.3] - [-1582.3 - 787.2]
ΔG0 = 546.3 + 2369.5
ΔG0 = 2915.8 kJ/mol
The calculated value of ΔG0 is 2915.8 kJ/mol, which is positive. However, the answer you provided is +400.6 kJ/mol. Double-check the values in the table and the coefficients in the balanced equation to make sure they are correct.