Chemistry Reference
In-Depth Information
values, the gas in the growing bubbles is continuously saturated with vapor of the
liquid (or partially decomposed liquid). As opposed to evaporation from the liquid
surface, the process rate is determined by the volumetric reaction kinetics and by
the humidity of the gas produced rather than by outward mass- and heat-transfer
parameters. In other words, the evaporation is a volumetric process. The chemical
reaction rate is significantly influenced by the evaporation since it occurs with heat
absorption and results in a reduction in the amount of reagent.
The situation where evaporation from the liquid surface prevails over the vol-
umetric process (for example, in very slow reactions in a thin liquid layer) is not
considered here, since it is not characteristic of explosion.
10.2 Kinetics of Thermal Decomposition of Volatile Explosives
Let us consider the kinetics of the reduction in the weight of a liquid under the
conditions of a thermogravimetric experiment at constant temperature. Under quasi-
stationary conditions, the total rate of the weight reduction (
dm
/
dt
) can be ex-
pressed as the sum of the gas formation and evaporation rates (
dm
g
/
dt
and
dm
v
/
dt
,
respectively):
−
dm
dt
=
dm
g
dt
+
dm
v
dt
−
,
1 +
dm
v
dm
g
.
(10.1)
dm
dt
=
dm
g
dt
−
In standard kinetic experiments the process rate is considerably low, while the mass-
and heat-transfer rates inside the bubbles are extremely high due to their small size.
Taking these circumstances into account, one can assume the existence of thermo-
dynamic equilibrium inside a bubble. In this case,
dm
v
dm
g
=
μ
v
P
v
μ
g
P
g
.
Taking into consideration that
P
v
+
P
g
=
P
∞
and
P
v
=
P
0
exp(
−
L
/
R
T
), one obtains
dm
v
dm
g
=
μ
v
P
0
exp(
−
L
/
R
T
)
L
/
R
T
)
,
(10.2)
μ
g
P
∞
−
P
0
exp(
−
where
T
is the liquid temperature,
P
∞
is the pressure in the gas bubble (equal to the
ambient pressure),
L
is the latent evaporation heat,
P
0
is the preexponential factor in
the expression for the saturated vapor pressure of the original liquid,
μ
g
are
the molecular weights of original liquid vapor and the gas produced, respectively,
and
R
is the gas constant.
Substituting Eq. (10.2) into Eq. (10.1), one obtains
μ
v
and
1 +
μ
v
μ
g
.
dm
dt
=
dm
g
dt
P
0
exp(
−
L
/
R
T
)
−
(10.3)
P
∞
−
P
0
exp(
−
L
/
R
T
)
