Page 236 - Physical chemistry understanding our chemical world

QUANTITATIVE EFFECTS OF PRESSURE AND TEMPERATURE CHANGE 203

Aside

Why does food cook faster at higher pressures?

The process of cooking involves a complicated series of chemical reactions, each of
which proceeds with a rate constant of k. When boiling an egg, for example, the rate-
limiting process is denaturation of the proteins from which albumen is made. Such
−1
denaturation has an activation energy E a of about 40 kJ mol .
The rate constant of reaction varies with tempera-
ture, with k increasing as the temperature increases. k We consider the Arrhe-
is a function of T according to the well-known Arrhe- nius equation in appro-
nius equation: priate detail in Chap-
ter 8.

k at T 2 E a 1 1
ln =− − (5.7)
k at T 1 R T 2 T 1
We saw in Worked Example 5.2 how the temperature of the boiling water increases
◦
◦
from 100 C to 147 C in a pressure cooker. A simple calculation with the Arrhenius
equation (Equation (5.7)) shows that the rate constant of cooking increases by a little
over fourfold at the higher temperature inside a pressure cooker.
◦
Boiling an egg takes about 4 min at 100 C, so boiling an egg in a pressure cooker
takes about 1 min.

Justification Box 5.2

The Clapeyron equation, Equation (5.1), yields a quantitative description of a phase
boundary on a phase diagram. Equation (5.1) works quite well for the liquid–solid
phase boundary, but if the equilibrium is boiling or sublimation – both of which involve
a gaseous phase – then the Clapeyron equation is a poor predictor.
For simplicity, we will suppose the phase change is the boiling of a liquid: liquid →
gas. We must make three assumptions if we are to derive a variant that can accommodate
the large changes in the volume of a gas:

Assumption 1: we assume the enthalpy of the phase change is independent of tem-
perature and pressure. This assumption is good over limited ranges of both p and T ,
although note how the Kirchhoff equation (Equation (3.19)) quantifies changes in H.

Assumption 2: we assume the gas is perfect, i.e. it obeys the ideal-gas equation,
Equation (1.13), so
pV = nRT or pV m = RT

where V m is the molar volume of the gas.