4. Calculate the Gibbs’ free energy change for the reaction
Mg(s) + H2O(g) → MgO(s) + H2(g)
carried out at 100ºC given the data below. Under these conditions is this
reaction product favoured?
ΔHºformation (kJ mol^-1)
MgO(g) -602
H2O(g) -242
H2(g) 0
Mg(s) 0
Sºformation (J K^-1 mol^-1)
MgO(g) 26.9
H2O(g) 188.7
H2(g) 130.6
Mg(s) 32.7
dHrxn = (n*dHproducts) - (n*dHreactants)
dSrxn = (n*dSproducts) - (n*dSreactants)
Then dGrxn = dHrxn - TdSrxn
If dGrxn is - the products are favored.
Show your work if you gets stuck.
To calculate the Gibbs' free energy change for the reaction, you will need to use the equation:
ΔGº = ΔHº - TΔSº
where:
ΔGº = Gibbs' free energy change
ΔHº = standard enthalpy change
T = temperature (in Kelvin)
ΔSº = standard entropy change
Given data:
ΔHº formation (kj mol^-1):
MgO(g) = -602
H2O(g) = -242
H2(g) = 0
Mg(s) = 0
Sº formation (J K^-1 mol^-1):
MgO(g) = 26.9
H2O(g) = 188.7
H2(g) = 130.6
Mg(s) = 32.7
First, you need to calculate the standard enthalpy change (ΔHº) for the reaction:
ΔHº = Σ ΔHº products - Σ ΔHº reactants
ΔHº = [ΔHº formation (MgO(s)) + ΔHº formation (H2(g))] - [ΔHº formation (Mg(s)) + ΔHº formation (H2O(g))]
ΔHº = [-602 + 0] - [0 -242]
ΔHº = -602 + 242
ΔHº = -360 kJ mol^-1
Next, calculate the standard entropy change (ΔSº) for the reaction:
ΔSº = Σ ΔSº products - Σ ΔSº reactants
ΔSº = [Sº formation (MgO(s)) + Sº formation (H2(g))] - [Sº formation (Mg(s)) + Sº formation (H2O(g))]
ΔSº = [26.9 + 130.6] - [32.7 + 188.7]
ΔSº = 157.5 - 221.4
ΔSº = -63.9 J K^-1 mol^-1
Lastly, substitute the values into the Gibbs' free energy equation:
ΔGº = ΔHº - TΔSº
Given T = 100ºC = 373 K,
ΔGº = -360 - (373 * -63.9)
ΔGº = -360 + 24016.7
ΔGº = 23656.7 J mol^-1
The Gibbs' free energy change for the reaction is 23656.7 J mol^-1.
To determine if the reaction is product-favored under these conditions, check the sign of ΔGº. If ΔGº is negative, the reaction is favored to form products. In this case, ΔGº is positive, so the reaction is not favored at 100ºC.
To calculate the Gibbs' free energy change (ΔGº) for the given reaction, we can use the equation:
ΔGº = ΔHº - TΔSº
where:
ΔHº is the enthalpy change,
ΔSº is the entropy change,
T is the temperature (in Kelvin).
Given data:
ΔHºformation (kJ mol^-1)
MgO(g) -602
H2O(g) -242
H2(g) 0
Mg(s) 0
Sºformation (J K^-1 mol^-1)
MgO(g) 26.9
H2O(g) 188.7
H2(g) 130.6
Mg(s) 32.7
We need to convert the entropy values from J K^-1 mol^-1 to kJ K^-1 mol^-1. So, divide them by 1000.
ΔSºformation (kJ K^-1 mol^-1)
MgO(g) 0.0269
H2O(g) 0.1887
H2(g) 0.1306
Mg(s) 0.0327
Now, we substitute the values into the equation:
ΔGº = ΔHº - TΔSº
Let's assume the temperature (T) is 100ºC. We need to convert it to Kelvin by adding 273.15.
T = 100ºC + 273.15 = 373.15 K
Now, let's calculate the ΔGº:
For the reactants:
ΔHº (Mg(s)) = 0 kJ mol^-1
ΔSº (Mg(s)) = 0.0327 kJ K^-1 mol^-1
For the products:
ΔHº (MgO(s)) = -602 kJ mol^-1
ΔSº (MgO(s)) = 0.0269 kJ K^-1 mol^-1
ΔHº (H2(g)) = 0 kJ mol^-1
ΔSº (H2(g)) = 0.1306 kJ K^-1 mol^-1
Now, let's substitute the values into the equation:
ΔGº = [ΔHº (MgO(s)) + ΔHº (H2(g))] - T[ΔSº (MgO(s)) + ΔSº (H2(g)) - ΔSº (Mg(s))]
ΔGº = [-602 + 0] - 373.15 * [0.0269 + 0.1306 - 0.0327]
ΔGº = -602 - 373.15 * [0.1248]
ΔGº = -602 - 46.45
ΔGº ≈ -648.45 kJ mol^-1
Since the value of ΔGº is negative (-648.45 kJ mol^-1), the reaction is