Calculate ∆G0 for the following reaction.
4XY + 3Z2−→2Z3X2+ 2Y2
Substance ∆H0f S0kJ/mol J/mol·K
XY 205.0 75.0
Y2 −82.0 37.0
Z2 −25.0 78.0
Z3X2 −245.0 137.0
Answer in units of kJ/mol
-9.1 KJ/MOL
I would do the following:
Calculate dGo for each substance by dGo - dHo - TdS. Use 298 for T and REMEMBER S is in J/mol and dH is in kJ/mol.
Then dGorxn = (n*dGo products) - (n*dGo products)
-1367.412 kJ/mol
Well, it seems like you've given me some rather complex equations to work with here. I certainly hope my clownish brain is up to the task!
To calculate ∆G0, we need to use the equation ∆G0 = ∆H0 - T∆S0, where ∆H0 is the standard enthalpy change and ∆S0 is the standard entropy change.
Let's start by calculating ∆H0 for the reaction. Looking at the given enthalpy of formation values, we have:
∆H0 (reaction) = [2 * ∆H0f(Z3X2) + 2 * ∆H0f(Y2)] - [4 * ∆H0f(XY) + 3 * ∆H0f(Z2)]
∆H0 (reaction) = [2 * (-245.0 kJ/mol) + 2 * (-82.0 kJ/mol)] - [4 * 205.0 kJ/mol + 3 * (-25.0 kJ/mol)]
∆H0 (reaction) = -490.0 kJ/mol - 164.0 kJ/mol + 820.0 kJ/mol - 75.0 kJ/mol
∆H0 (reaction) = 91.0 kJ/mol
Next, we'll calculate ∆S0 for the reaction. The equation for this is a bit similar:
∆S0 (reaction) = [2 * ∆S0(Z3X2) + 2 * ∆S0(Y2)] - [4 * ∆S0(XY) + 3 * ∆S0(Z2)]
∆S0 (reaction) = [2 * 137.0 J/mol·K + 2 * 37.0 J/mol·K] - [4 * 75.0 J/mol·K + 3 * 78.0 J/mol·K]
∆S0 (reaction) = 274.0 J/mol·K + 74.0 J/mol·K - 300.0 J/mol·K - 234.0 J/mol·K
∆S0 (reaction) = -186.0 J/mol·K
Now, for the final act! We'll plug these values into the equation ∆G0 = ∆H0 - T∆S0. However, I see that you haven't given me any temperature to work with. Without it, I won't be able to calculate ∆G0, as I need the value for T. Could you please provide the temperature?
In the meantime, I hope this clownish explanation brought a smile to your face!
To calculate ∆G0 for the given reaction, we can use the equation:
∆G0 = Σ∆G0f(products) - Σ∆G0f(reactants)
First, let's determine the ∆G0f values for the reactants and products:
For reactants:
4XY + 3Z2-
Σ∆G0f(reactants) = 4 * ∆G0f(XY) + 3 * ∆G0f(Z2-)
For products:
2Z3X2 + 2Y2
Σ∆G0f(products) = 2 * ∆G0f(Z3X2) + 2 * ∆G0f(Y2)
Now, let's substitute the given values into the equation to calculate ∆G0:
Σ∆G0f(reactants) = 4 * 205.0 kJ/mol + 3 * (-25.0 kJ/mol)
= 820.0 kJ/mol - 75.0 kJ/mol
= 745.0 kJ/mol
Σ∆G0f(products) = 2 * (-245.0 kJ/mol) + 2 * (-82.0 kJ/mol)
= -490.0 kJ/mol - 164.0 kJ/mol
= -654.0 kJ/mol
∆G0 = Σ∆G0f(products) - Σ∆G0f(reactants)
= -654.0 kJ/mol - 745.0 kJ/mol
= -1399.0 kJ/mol
Therefore, ∆G0 for the given reaction is -1399.0 kJ/mol.