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In this lesson we will review Hess’s law and how it is used in calculations. We will also solve for Gibbs Free Energy.
Hess’s law states that in a reaction the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.
To determine the change in enthalpy, we may rearrange the given equations to obtain the desired equation, and then add the corresponding changes in enthalpy to determine the enthalpy for the overall reaction.
Given the following equation: B → E + 2C, calculate ∆H.
The intermediate steps and their respective enthalpies are given below
| ∆H (kJ/mol) | |
| (1/2)A → B | 150 |
| 3B → 2C + D | -125 |
| E + A → D | 350 |
To obtain the third given equation, select the substance that only appears in one of the intermediate equations, in this case E. The desired equation has E on the reactants side, but we want it to be on the products side, so reversing the third given equation we get:
D → E + A, ∆H = -350 kJ/mol
Then select another substance that only appears once in the given equations and that is in the desired equation, in
this case C. 2C is desired to be on the products side in the final equation and it is in the second given equation,
so the second step stays the same.
3B → 2C + D, ∆H=-125.
Next, there is only B remaining in the desired equation. 3B is already a reactant, but only B is desired, so 2B
must be on the products side to cancel out 2B from the reactant side. The first equation with B as a product must be
multiplied by two to cancel out 2B from the reactants side. Remember also to multiply the change in enthalpy by 2
because the amount of products and reactants is doubled
A → 2B, ∆H = 300
Next, add the three reactions and the corresponding changes in enthalpy, then cancel as necessary.
| ∆H (kJ/mol) | |
|---|---|
| D → E + A | -350 |
| 3B → 2C + D | -125 |
| A → 2B | 300 |
| D + 3B + A → E + A + 2C + D + 2B | -175 kJ/mol |
Now, cancel the terms in the same amount that appear on both sides of the equation. One D is on both sides and cancels. Two B’s are on either side of the arrow and cancel as does the one A.
∆H for B → E + 2C is -175 kJ/mol
Use Hess’s law to find the change in enthalpy at 25°C for the following equation:
CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)
Given:
| ∆H (kJ) | |
| Ca(s) + 2C(graphite) → CaC2(s) | -62.8 |
| Ca(s) + (1/2)O2(g) → CaO(s) | -635.5 |
| CaO(s) + H2O(l) → Ca(OH)2(aq) | -653.1 |
| C2H2(g) + (5/2)O2(g) → 2CO2(g)+ H2O(l) |
-1300 |
| C(graphite) + O2(g) → CO2(g) | -393.51 |
The correct answer is B. Solve the problem as follows: CaC2 is desired on the reactants side. The only equation with CaC2 is the first one, but to get calcium carbide as a reactant the equation must be reversed. This also causes the enthalpy to become positive.
CaC2(s) → 2C(graphite) + Ca(s), ΔH = +62.8kJ
Skip water since it appears in multiple places in the given equations. Calcium hydroxide is a desired product and the only reaction with the substance is the third equation, where it is a product. This reaction will stay the same.
CaO(s) + H2O(l) → Ca(OH)2(aq), ΔH = –653.1 kJ
Acetylene is a desired product. It appears as a reactant in the fourth equation, so that equation will need to be reversed as well as the sign of the enthalpy.
2CO2 + H2O(l) → C2H2(g) + (5/2)O2(g), ΔH = +1300 kJ
Next, two carbons in the form of graphite are currently products due to reversing the first equation. Since carbon is not a desired product, two carbons must be reactants to cancel the two that appear as products. Hence, the fifth equation needs to be multiplied by two, as will its enthalpy.
2C(graphite) + 2O2(g) → 2CO2(g), ΔH = –787.02 kJ
Taking a look at the second equation, if left as is the equation will cancel the calcium and calcium oxide from the other equations that are not a desired product. This reaction stays as is.
Ca(s) + (1/2)O2(g) → CaO(s), ΔH = –635.5 kJ
Add the five equations, their respective enthalpies and cancel as necessary.
After canceling, we get the following equation:
CaC2(s) + 2H2O(l) → Ca(OH)2(aq) +
C2H2(g) , ΔH = -7.13 x 102 kJ
Free energy is a thermodynamic function that is equal in value to the enthalpy minus the Kelvin temperature multiplied by the entropy.
The change in Gibbs Free Energy can be determined using the equation
ΔG =ΔH – TΔS
G is free energy, H is enthalpy, T is temperature in Kelvin, and S is entropy. The values for ΔG and ΔH are measured in kJ, ΔS is measured in joules, so the value must be divided by 1000 to get kilojoules.
To solve, we must find the total enthalpy of the products minus the enthalpy of the reactants. We can use Hess’s Law to solve for changes in entropy and enthalpy. Thus:
ΔG =ΣGproducts − ΣGreactants
| ΔG | ΔH | ΔS | Reaction |
|---|---|---|---|
| (-) | (-) | (+) | The reaction is always spontaneous |
| (+ or -) | (+) | (+) | The reaction is spontaneous at high temperatures |
| (+ or -) | (-) | (-) | The reaction is spontaneous at low temperatures |
| (+) | (+) | (-) | The reaction is never spontaneous |
When ΔH = TΔS, the value for ΔG = 0 and the reaction is at equilibrium.
Find Gibbs free energy for the following reaction at 298 K.
Fe3O4(s, magnetite) + Al(s) → Al2O3(s) + Fe(s)
To find the values for the substances, look at a chart of thermodynamic data.
| Substance | ΔH(kJ/mol) | S(J/mol·K) |
|---|---|---|
| Al2O3(s) | -1676 | 51 |
| Fe(s) | 0 | 27 |
| Fe3O4(s, magnetite) | -1117 | 146 |
| Al(s) | 0 | 28 |
First, remember to balance the equation:
3Fe3O4(s, magnetite) + 8Al(s) → 4Al2O3(s) + 9Fe(s)
ΔH = (ΣHproducts − ΣHreactants)
ΔH = [4 mol(-1676 kJ/mol) – 9 mol(0 kJ/mol)] – [8 mol(0 kJ/mol) + 3 mol(-1117 kJ/mol)]
ΔH = -3353 kJ
ΔS = 
ΔS = -0.21 kJ/K
ΔG =ΔH – TΔS
ΔG =-3353 kJ – (298 K)(-0.21 kJ/K)
ΔG = -3300 kJ
Even though the value of the entropy change is negative the temperature is low enough that the negative enthalpy term still allows the free energy to be negative. Thus the reaction is spontaneous.
Determine the free energy and spontaneity for the following reaction that occurs at 333 K.
KClO3(s) → KCl(s) + O2(g)
| Substance | ΔH(kJ/mol) | S(J/mol·K) |
|---|---|---|
| KClO3(s) | -391 | 143 |
| KCl(s) |
-436
|
83
|
|
KCl(aq)
|
-450
|
205
|
|
O2(g)
|
0
|
28
|
|
O2(aq)
|
-12
|
111
|
The correct answer is B.
2KClO3(s) → 2KCl(s)+ 3O2(g)
ΔH = (ΣHproducts − ΣHreactants)
ΔH = [3 mol(0 kJ/mol) + 2 mol(-436 kJ/mol)] – [2 mol(-391 kJ/mol)]
ΔH = -90 kJ
ΔS =
ΔS = [3 mol(28 kJ/mol) + 2 mol(83 kJ/mol)] – [2 mol(143 kJ/mol)]
ΔS = -36 kJ
ΔG = -90 kJ – (333 K)(-0.036 kJ)
ΔG = -78 kJ
The reaction is spontaneous.
Determine Gibbs free energy and the spontaneity for the following reaction at 34.5°C.
NH3(g) + HCl(aq) → NH4Cl(aq)
|
Substance
|
ΔH(kJ/mol)
|
ΔS(J/mol·K)
|
|
NH3(g)
|
–46
|
193
|
|
NH3(aq)
|
–80
|
111
|
|
HCl(aq)
|
–167
|
57
|
|
HCl(g)
|
–92
|
187
|
|
NH4Cl(aq)
|
–300
|
170
|
|
NH4Cl(s)
|
–314
|
96
|
The correct answer is B.
NH3(g) + HCl(aq) → NH4Cl(aq)
ΔH = (ΣHproducts − ΣHreactants)
ΔH = [1 mol(-300 kJ/mol)] – [1 mol (-46 kJ/mol) + 1 mol (-167kJ/mol)]
ΔH = -87 kJ
ΔS =
ΔS = 
ΔS = -0.080 kJ/K
ΔG = -87 kJ – (307.65 K)(-0.080 kJ/K)
ΔG = -62 kJ
The reaction is spontaneous at low temperatures.