Enthalpy of Transition
213
the molecules or ions and permit them to move more independently than they
could in their former state; the new state with its added energy always has less
molecular order. For example, liquid water at 0°C is a less-ordered state than
crystalline water at 0°C, and water vapor at 100°C is chaotic in its molecular
organization compared to liquid water at 100°C.
These statements are made more precise and quantitative in the following
way. It is said that, at a given temperature and pressure, the molecules of each
substance in "state 1" (say, liquid) have a heat content of//,, whereas in "state
2" (say, vapor) they have a heat content of H 2 . The "heat of transition" (in this
case, vaporization) is simply "the change in heat content" (A// T ):
A//T = //2 - //,
To avoid the use of the ambiguous term "heat" in connection with "heat
content," it is customary to use the term enthalpy. At a given temperature and
pressure, every substance possesses a characteristic amount of enthalpy (//),
and the heat changes associated with chemical and physical changes at constant
pressure are called changes in enthalpy (A//); A// T is the enthalpy of transition.
Two common enthalpies of transition are A//f = 1435 cal/mole for the enthalpy
effusion (melting) of ice at 0°C, and A// x = 9713 cal/mole for the enthalpy of
vaporization of water at 100°C.
Energy also is involved in transitions from one allotropic form to another, or
from one crystal form to another. To change a mole of red phosphorus to yellow
phosphorus, we must supply 4.22 kilocalories (A// T = +4.22 kcal/mole), and
when 1 mole of yellow silicon disulfide changes to white silicon disulfide, 3.11
kcal is liberated (A// T = -3.11 kcal/mole).
In the following problem we apply the principles involved in both specific
heat and heats of transition.
PROBLEM:
What is the resulting temperature if 36.0 grams of ice at 0°C are put into 200 g of
H 2 O at 25.0°C, contained in the calibrated calorimeter used in the preceding
problems?
SOLUTION:
The energy required to melt the ice is supplied by the water and the calorimeter
walls which, as a result, are cooled. Let T be the final temperature.
Calories needed to melt ice at 0°C =
'
= 2870 cal
(18.0 g/mole)
(
cal \
1 — - UrC) = 36.0T cal
g C/
g
Total calories needed = 2870 + 36.07
213
the molecules or ions and permit them to move more independently than they
could in their former state; the new state with its added energy always has less
molecular order. For example, liquid water at 0°C is a less-ordered state than
crystalline water at 0°C, and water vapor at 100°C is chaotic in its molecular
organization compared to liquid water at 100°C.
These statements are made more precise and quantitative in the following
way. It is said that, at a given temperature and pressure, the molecules of each
substance in "state 1" (say, liquid) have a heat content of//,, whereas in "state
2" (say, vapor) they have a heat content of H 2 . The "heat of transition" (in this
case, vaporization) is simply "the change in heat content" (A// T ):
A//T = //2 - //,
To avoid the use of the ambiguous term "heat" in connection with "heat
content," it is customary to use the term enthalpy. At a given temperature and
pressure, every substance possesses a characteristic amount of enthalpy (//),
and the heat changes associated with chemical and physical changes at constant
pressure are called changes in enthalpy (A//); A// T is the enthalpy of transition.
Two common enthalpies of transition are A//f = 1435 cal/mole for the enthalpy
effusion (melting) of ice at 0°C, and A// x = 9713 cal/mole for the enthalpy of
vaporization of water at 100°C.
Energy also is involved in transitions from one allotropic form to another, or
from one crystal form to another. To change a mole of red phosphorus to yellow
phosphorus, we must supply 4.22 kilocalories (A// T = +4.22 kcal/mole), and
when 1 mole of yellow silicon disulfide changes to white silicon disulfide, 3.11
kcal is liberated (A// T = -3.11 kcal/mole).
In the following problem we apply the principles involved in both specific
heat and heats of transition.
PROBLEM:
What is the resulting temperature if 36.0 grams of ice at 0°C are put into 200 g of
H 2 O at 25.0°C, contained in the calibrated calorimeter used in the preceding
problems?
SOLUTION:
The energy required to melt the ice is supplied by the water and the calorimeter
walls which, as a result, are cooled. Let T be the final temperature.
Calories needed to melt ice at 0°C =
'
= 2870 cal
(18.0 g/mole)
(
cal \
1 — - UrC) = 36.0T cal
g C/
g
Total calories needed = 2870 + 36.07