Enthalpy and Chemical Reactions

Problem 1:
When Zinc is added to an acid solution, such as hydrochloric acid, the zinc
reacts and hydrogen gas is evolved. The reaction can be described by the
equation.
Zn (s) + 2H+ (aq) --à Zn 2+ (aq) + H 2 (g)
The solution becomes warm, showing that the reaction evolves heat; i.e., it is
an exothermic reaction. Reconcile these experimental results given below,
and in so doing calculate values for the change in the internal energy and the
enthalpy of the system.
(a) If the reaction were carried out in a constant-pressure calorimeter, as in a
simple Styrofoam cup experiment, it would be found that the thermal
surroundings gain 153.89 kJ per mol of zinc that reacts.
(b) If the reaction were carried out in a constant-volume "bomb calorimeter,"
it would be found that the thermal surroundings gain 156.37 kJ per mole
of zinc reacted.
(a)
For constant-pressure processes:
∆H = - ∆U therm = - (+153.89 kJmol -1 ) = -153.89 kJmol -1
Since
∆H
= ∆U + ∆ν g (RT)
We can write
∆U = ∆H - ∆ν g (RT)
= (-153.89 kJ mol -1 ) - (+1) (8.314 JK -1 mol -1 ) (298 K)

= (-153.89 kJ mol -1 ) - (+1) (8.314 JK -1 mol -1 )(298 K)
= (-153.89 kJ mol -1 ) - (2477.572 Jmol -1 )
= (-153.89 kJ mol -1 ) - (2.48 kJmol -1 )
= -156.37 kJ mol -1
(b)
For constant-volume processes:
∆U = - ∆U therm = = - (+156.37 kJmol -1 ) = -156.37 kJmol -1
Since
∆H
= ∆U + ∆ν g (RT)
= (-156.37 kJ mol -1 ) + (+1) (8.314 JK -1 mol -1 ) (298 K)
= (-156.37 kJ mol -1 ) + (+1) (8.314 JK -1 mol -1 )(298 K)
= (-156.37 kJ mol -1 ) + (2477.572 Jmol -1 )
= (-156.37 kJ mol -1 ) + (2.48 kJmol -1 )
= -153.89 kJ mol -1
Problem 2:
The cylinder containing 1 mol of liquid water at 100 0 C is heated until the
liquid is converted to vapor. Th cylinder is fitted with a piston which applies
a pressure of 1 bar. (a) How much energy is transferred to mechanical
surroundings (b) If the enthalpy of vaporization of water is 40670 J, what is
the change in internal energy

(a) ∆U mech
= P ext ∆V
= P ext V 2
= P ext nRT /P
= nRT
= (1 mol) (8.314 J mol -1 k -1 ) (373 K)
= 3101.12 J
(b) ∆H = ∆U + ∆U mech
∆U = ∆H - ∆U mech
= 40670 J - 3101 J
= 37568.88 J = 37.57 kJ
Enthalpy and Chemical reactions:
1. The energy change that accompanies a chemical reaction depends upon
whether each of the reagents is solid, liquid or gas.
This is specified by adding s, l or g after the formula.
2. Energies are written based on the molar amounts of the reagents and not
on 1 mol of any one of the reagents.
Reaction in which 1 mol of H 2 O is formed.

H 2 (g) + (1/2)O 2 (g) ---à
H 2 O (l)
∆H = -285.83 kJ mol -1
and
Reaction in which 2 mole of H 2 O is formed
2H 2 (g) + O 2 (g) ---à 2H 2 O (l)
∆H = -571.66 kJ mol -1
3. Combustion reactions are written as the combustion for 1 mol of the
reactants.
C 6 H 6 (g) + (15/2)O 2 (g) ---à 6CO 2 (g) + 3H 2 O (l)
∆H = -3301.48 kJ mol -1
4. If reaction is carried out at standard pressure (1 bar) and a temperature of
250C (i.e. 298.15 K).
These conditions are indicated by a subscript 298 (suggesting a temperature
value) and a degree sign implying standard pressure.
H 2 (g) + (1/2)O 2 (g) ---à
H 2 O (l)
∆H 0 298 = -285.83 kJ mol -1
Indirect determination of Enthalpy change in chemical reactions:
Many reactions are not suitable for direct calorimetric study. In such cases,
the internal energy change or enthalpy change can be obtained by an indirect
method. This procedure was developed by a scientist Hess in 1840 and
hence called Hess' Law of Heat Summation.

This is based on the First Law of Thermodynamics i.e. energy is a property
and that changes in the energy depend only on the initial and final states and
not on the path between these states.
For e.g. conversion of carbon in graphite form to its diamond form,
C (graphite)
----à C (diamond)
Since this reaction can be made to occur, it is not suitable for direct
calorimetric measurement.
But, the combustion of both graphite and diamond can be carried out in a
calorimeter.
These reactions and their enthalpy changes for the combustion reactions are
written as:
C (graphite) + O 2 (g) ---à
CO 2 (g)
∆H 0 298 = -393.51 kJ mol -1
C (diamond) + O 2 (g) ---à
CO 2 (g)
∆H 0 298 = -395.40 kJ mol -1
The graphical display for the enthalpy change for both these reactions is
shown as follows:
Refer to Figure 3.4

On the basis of this diagram, we can write
C (graphite) --à C (diamond)
∆H 0 298 = +1.89 kJ mol -1
The another way of obtaining the result is by calcualtion.
1. Addition:
The second equation is turn around and equations are added and common
products are cancelled.
C (graphite) + O 2 (g) ---à
CO 2 (g)
∆H 0 298 = -393.51 kJ mol -1
CO 2 (g)
---à C (diamond) + O 2 (g)
∆H 0 298 = +395.40 kJ mol -1
C (graphite) --à C (diamond)
∆H 0 298 = - 393.51 kJ mol -1 + (+395.40 kJ mol -1 )
= + 1.89 kJ mol -1
2.Substraction:
The second equation is subtracted from first equation.

C (graphite) + O 2 (g) ---à
CO 2 (g)
∆H 0 298 = -393.51 kJ mol -1
C (diamond) + O 2 (g) ---à CO 2 (g)
∆H 0 298 = +395.40 kJ mol -1
C (graphite) --à C (diamond)
∆H 0 298 = - 393.51 kJ mol -1 - (- 395.40 kJ mol -1 )
= + 1.89 kJ mol -1
Problem 3:
Calculate the enthalpy change for the formation of methane from its
elements according to the equation:
C (graphite) - 2H 2 (g) ---à CH 4 (g)
The values for enthalpy changes for combustion of C(graphite), H 2 (g) and
CH 4 (g) in units of kilojoules per mole are -393.51, -285.84 and -890.35.
The combustion reactions for C(graphite), H 2 (g) and CH 4 (g) are written as
follows:
(A)
C (graphite) + O 2 (g) ---à CO 2 (g)
∆H 0 298 = -393.51 kJ mol -1

(b)
H 2 (g) + (1/2)O 2 (g) ---à
H 2 O (l)
∆H 0 298 = -285.83 kJ mol -1
(c)
CH 4 (g) + 2O 2 (g) ---à CO 2 (g) + 2H 2 O (l)
∆H 0 298 = -890.35 kJ mol -1
Then
(a) + 2(b) - (c):
C (graphite) - 2H 2 (g) ---à CH 4 (g)
∆H 0 298 = -74.84 kJ mol -1