208
Thermochemistry
The insulation surrounding the calorimeter minimizes heat loss or gain
through the walls. The cover (C) supports the electrical heater (H); it has two
holes, one to permit the insertion of a sensitive differential thermometer (T), and
one through which reactants can be added. The solution is stirred by a Tefloncovered magnet (M) that is rotated by the motor-driven magnet (M'). The
electrical heating is controlled and measured as follows. The double-pole
switch (S) controls two things simultaneously: the timer (W), and the battery
(B), which supplies a constant current to the immersion heater (H). The current
(/) that flows through the heater is read from the ammeter (A), and the duration
(t) of time that the current flows is read from the timer. The resistance (R) of
the heater is known from a separate measurement. The following problems are
based on the calorimeter shown in Figure 14-1.
PROBLEM:
A calorimeter contains 200 ml of 0.100 M NaCl solution at room temperature
(25.038°C according to the thermometer). With the magnetic stirrer going, exactly
10.00 ml of 1.000 M AgNO 3 solution (also at room temperature) is added dropwise
through the porthole of the calorimeter. The temperature, as indicated by the
thermometer, rises to 25.662°C. A current of 0.700 ampere is then passed through
the electrical heater (whose resistance is 6.50 ohms) for a period of 4 min 5 sec,
and the temperature rises by 0.742°C. Calculate the number of calories produced
when one mole of AgCl precipitates from aqueous solution.
SOLUTION:
The electrical heater and thermometer tell you how many calories it takes to raise
the reaction mixture by 1°C:
(0.700 amp)
2 (6.50 ohm)(245 sec)
4.184^
cal
energy
186.5 cal
0.742°C
" 0.742°C
The chemical reaction caused the temperature of the same mixture to rise by
0.624°C, corresponding to an energy release of
(»' f) -
energy from reaction = (0.624°C) ( 251 — j = 157 cal
You will note that there was added (0.200 liter) (0.100 mole/liter) = 0.0200 mole
NaCl and (0.0100 liter) (1.00 mole/liter) = 0.0100 mole AgNO 3 . Only half of the
NaCl is used, and just 0.0100 mole of AgCl is formed. It is the formation of this
0.0100 mole of AgCl that produces the 157 cal, so
Thermochemistry
The insulation surrounding the calorimeter minimizes heat loss or gain
through the walls. The cover (C) supports the electrical heater (H); it has two
holes, one to permit the insertion of a sensitive differential thermometer (T), and
one through which reactants can be added. The solution is stirred by a Tefloncovered magnet (M) that is rotated by the motor-driven magnet (M'). The
electrical heating is controlled and measured as follows. The double-pole
switch (S) controls two things simultaneously: the timer (W), and the battery
(B), which supplies a constant current to the immersion heater (H). The current
(/) that flows through the heater is read from the ammeter (A), and the duration
(t) of time that the current flows is read from the timer. The resistance (R) of
the heater is known from a separate measurement. The following problems are
based on the calorimeter shown in Figure 14-1.
PROBLEM:
A calorimeter contains 200 ml of 0.100 M NaCl solution at room temperature
(25.038°C according to the thermometer). With the magnetic stirrer going, exactly
10.00 ml of 1.000 M AgNO 3 solution (also at room temperature) is added dropwise
through the porthole of the calorimeter. The temperature, as indicated by the
thermometer, rises to 25.662°C. A current of 0.700 ampere is then passed through
the electrical heater (whose resistance is 6.50 ohms) for a period of 4 min 5 sec,
and the temperature rises by 0.742°C. Calculate the number of calories produced
when one mole of AgCl precipitates from aqueous solution.
SOLUTION:
The electrical heater and thermometer tell you how many calories it takes to raise
the reaction mixture by 1°C:
(0.700 amp)
2 (6.50 ohm)(245 sec)
4.184^
cal
energy
186.5 cal
0.742°C
" 0.742°C
The chemical reaction caused the temperature of the same mixture to rise by
0.624°C, corresponding to an energy release of
(»' f) -
energy from reaction = (0.624°C) ( 251 — j = 157 cal
You will note that there was added (0.200 liter) (0.100 mole/liter) = 0.0200 mole
NaCl and (0.0100 liter) (1.00 mole/liter) = 0.0100 mole AgNO 3 . Only half of the
NaCl is used, and just 0.0100 mole of AgCl is formed. It is the formation of this
0.0100 mole of AgCl that produces the 157 cal, so