Thermochemical
Equations

(See
^{nd} ed., Sections 6.5 & 6.6, pp.
233-240)

The
heat flow for a reaction at constant pressure, q_{p}, is called
enthalpy, ΔH.

Exothermic
reactions have negative enthalpy values (-ΔH).

Endothermic
reactions have positive enthalpy values (+ΔH).

An
equation which shows both mass and heat relationships between products and
reactants is called a thermochemical equation.

2
H_{2(g)} + O_{2(g)} ----> 2 H_{2}O_{(l)}
ΔH = -571.6 kJ

The
magnitude of ΔH is directly proportional to the amount of reactants or
products.

A
+ 2 B ----> C
ΔH = -100 kJ

1/2
A + B ----> 1/2 C
ΔH = -50 kJ

ΔH
for a reaction is equal in magnitude but opposite in sign for the reverse
reaction.

A
+ 2 B ----> C
ΔH = -100 kJ

C
----> A + 2 B
ΔH = +100 kJ

__Example
1
__

Based
upon the thermochemical equation given, calculate the heat associated with the
decomposition of 1.15 g of
NO_{2(g)}.

N_{2(g)}
+ 2 O_{2(g)} ----> 2 NO_{2(g)}
ΔH = +67.6 kJ

__Example
2
__Using
the following thermochemical equation, calculate how much heat is associated
with the decomposition of 4.00 moles of NH

NH_{3}(g)
+ HCl(g) ---> NH_{4}Cl(s) ΔH =
-176 kJ

__Example
3
__What
quantity of heat is associated with the synthesis of 45.0 g of NH

4
NO(g) + 6 H_{2}O(l) ---> 4 NH_{3}(g) + 5 O_{2}(g)
ΔH = +1170 kJ

__Example
4__

Calculate the mass of ethane, C_{2}H_{6}, which must be burned
to produce 100 kJ of heat.

2
C_{2}H_{6}(g) + 7 O_{2}(g) ---> 4 CO_{2}(g) +
6 H_{2}O(l) ΔH = -3120 kJ

Hess’s
Law

(See
^{nd} ed., Section 6.9, pp.
246-247)

For
a chemical equation that can be written as the sum of two or more steps, the
enthalpy change for the overall equation equals the sum of the enthalpy changes
for the individual steps.
In other words, no matter how you go from given reactants to products
(whether in one step or several), the enthalpy change for the overall chemical
change is the same.

The
heat transferred in a given change is the same whether the change takes place in
a single step or in several steps.

ΔH_{rxn}
for a given reaction is the same whether that reaction takes place directly
through one step or via several different reactions.

S_{(s)} + O_{2 (g)} ---> SO_{2 (g)}
ΔH = -296 kJ

SO_{2 (g)} + 1/2 O_{2 (g)} ---> SO_{3 (g)}
ΔH = -98.9 kJ

____________________________________________

S_{(s)} + 1 1/2 O_{2 (g)} ---> SO_{3
(g)} ΔH = -394.9 kJ

In
order to use Hess's Law:

1)
If a rxn is reversed, the sign of ΔH is reversed.

S_{(s)} + O_{2 (g)} ---> SO_{2 (g)} ΔH =
-296 kJ

SO_{2 (g)} ---> S_{(s)}+ O_{2 (g)} ΔH =
+296 kJ

2)
If the coefficients in a balanced equation are multiplied by some number, the
value of ΔH must be multiplied by that same number.

S_{(s)} + O_{2 (g)} ---> SO_{2 (g)} ΔH = -296 kJ

2 S_{(s)} + 2 O_{2 (g)} ---> 2 SO_{2 (g)}
ΔH = (2)(-296 kJ) = -592 kJ

__Example
5__

Calculate ΔH for the vaporization of water from liquid, using the
thermochemical equations given.

H_{2}O_{(l)
}--->_{ }H_{2}O_{(g)}
ΔH = ?

a) H_{2 (g)} + 1/2 O_{2 (g)} --->H_{2}O_{(l)}
ΔH = -286 kJ

b) H_{2 (g)} + 1/2 O_{2 (g)} ---> H_{2}O_{(g)}
ΔH = -242 kJ

__Example
6__

Calculate ΔH for the following reaction using the thermochemical equations
given.

2
C_{(s)} + H_{2 (g)} ---> C_{2}H_{2 (g)}
ΔH = ?

a) C_{(s)} + O_{2 (g)} ---> CO_{2 (g)}
ΔH = -393.5 kJ

b) H_{2 (g)} + 1/2 O_{2 (g)} ---> H_{2}O_{(l)}
ΔH = -285.8 k J

c) 2 C_{2}H_{2 (g)} + 5 O_{2 (g)} ---> 4 CO_{2
(g)} + 2 H_{2}O_{(l) } ΔH
= -2598.8 kJ