Download Thermodynamics Part 3

Thermodynamics Part 3
Hess’s Law
Enthalpy (∆ H)
• Total energy content of a system
• Heat gained or lost when reactions occur
• Energy gained or lost when bonds break & form
Hess’s Law
• The law states that the energy change for any
chemical or physical process is independent of
the pathway or number of steps required to
complete the process provided that the final
and initial reaction conditions are the same. In
other words, an energy change is path
independent, only the initial and final states
being of importance. This path independence
is true for all state functions.
continued….
Hess's law allows the enthalpy change (ΔH) for a
reaction to be calculated even when it cannot be
measured directly. This is accomplished by
performing arithmetic operations on chemical
equations and known ΔH values. Chemical
equations may be multiplied or divided by a
whole number. When an equation is multiplied
by a constant, its ΔH must be multiplied by the
same number as well. If an equation is reversed,
ΔH for the reaction must also be reversed
….continued….
• Addition of chemical equations can lead to a net
equation. If enthalpy change is included for each
equation and added, the result will be the
enthalpy change for the net equation. If the net
enthalpy change is negative (ΔHnet < 0), the
reaction will be exothermic and is more likely to
be spontaneous; positive ΔH values correspond
to endothermic reactions.
….continued….
• Hess's Law says that enthalpy changes are
additive. Thus the ΔH for a single reaction can
be calculated from the difference between
theheat of formation of the products and the
heat of formation of the reactants. In
mathematical terms:
Example
Given:
•
•
•
•
B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g) (ΔH = 2035 kJ/mol)
H2O (l) → H2O (g) (ΔH = 44 kJ/mol)
H2 (g) + (1/2) O2 (g) → H2O (l) (ΔH = -286 kJ/mol)
2 B (s) + 3 H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol)
Find the ΔHf of:
2 B (s) + (3/2) O2 (g) → B2O3 (s)
Answer….
• After the multiplication and reversing of the equations (and
their enthalpy changes), the result is:
• B2H6 (g) + 3 O2 (g) → B2O3 (s) + 3 H2O (g) (ΔH = -2035 kJ/mol)
• 3 H2O (g) → 3 H2O (l) (ΔH = -132 kJ/mol)
• 3 H2O (l) → 3 H2 (g) + (3/2) O2 (g) (ΔH = 858 kJ/mol)
• 2 B (s) + 3 H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol)
Adding these equations and canceling out the common
terms on both sides, we get
2 B (s) + (3/2) O2 (g) → B2O3 (s) (ΔH = -1273 kJ/mol)
Hmmm…
OK….so what do
we need to
know?
Hess’s Law
• The total enthalpy change for a reaction is
equal to the sum of the enthalpy changes for
each of the small reactions
∆H
∆H
reaction
=
TOTAL =
∆H
formation prod
∆H
f (prod)
-
∆H
formation reactants
- ∆H
f (reactants)
Rules/Steps for Solving Hess’s Law
1. Write balance chemical equation
2. Find ∆ Hf for each reactant (using reference table)
and multiply ∆ H by the # moles of reactants
(stoichiometry)
3. Find ∆ Hf for each product (using reference table)
and multiply ∆ H by the # moles of reactants
(stoichiometry)
4. Plug into Hess’s Law equation
∆ HTOTAL = ∆ H f (prod) - ∆ H f (reactants)
(this will solve for kJ reaction/mole generic reaction)
5. Convert grams to moles from mass given
6. Multiply moles used (step 5) x ∆ HTOTAL per
stoichiometry
example #1
• Determine ∆ H for the combustion of 5 grams
of methane gas:
CH4 + 2 O2 →
CO2 + 2 H2O (g)
∆ HTOTAL = ∆ H f (prod) - ∆ H f (reactants)
∆ Hproducts = -393.5 + 2 (-242)
∆ Hreactants = -74.8 + 2 (0)
= - 877.5 kJ/mole
= - 74.8 kJ/mole
∆ HTOTAL = (- 877.5 kJ/mole) – (-74.8 kJ/mole) = - 803 kJ/mole
5 g CH4
1 mole CH4
= .3125 moles CH4
16 g CH4
= - 250.9 kJ
- 803 kJ
Overall
Reaction
1 mole CH4
Heat produced from 5 grams
of Methane
example #2
• Determine ∆ H for spontaneous reaction of
5 grams of Sodium metal with water.
2 Na + 2 H2O(l) → 2 NaOH + H2 (g)
∆ HTOTAL = ∆ H f (prod) - ∆ H f (reactants)
∆ Hproducts = 2 (- 427) + 2 (0)
∆ Hreactants = 2 (0) + 2 (- 286)
= - 854 kJ/mole
= - 572 kJ/mole
∆ HTOTAL = (- 854 kJ/mole) – (- 572 kJ/mole)
5 g Na
1 mole Na
= - 282 kJ/mole
= .217 moles Na - 282 kJ
23 g Na
= - 30.6 kJ
Overall
Reaction
2 mole Na
Heat produced from 5 grams
of Sodium
CHALLENGE #1
• Determine ∆ H for the following reaction
assuming 1,500 grams of Iron
2 Fe + 3 CO2 → Fe2O3
+3 CO
∆ HTOTAL = ∆ H f (prod) - ∆ H f (reactants)
∆ Hproducts = - 822 + 3 ( - 111)
∆ Hreactants = 2 (0) + 3 ( -393.5)
= - 1155 kJ/mole
= - 1180.5 kJ/mole
∆ HTOTAL = (- 1155 kJ/mole) – (- 1180.5 kJ/mole)
1500 g Fe 1 mole Fe
= 25.5 kJ/mole
= 26.79 moles Fe + 25.5 kJ
56 g Fe
= + 341. 5 kJ
Overall
Reaction
2 mole Fe
Heat absorbed during reaction of
1,500 grams of Iron
Challenge #2
• Determine ∆ H for the following reaction
assuming 1,500 grams of Iron are produced.
Fe2O3
+ 3 CO
→ 2 Fe + 3 CO2
∆ H = - 25.5 kJ/mole