Page 137 - Physical chemistry understanding our chemical world

104 ENERGY AND THE FIRST LAW OF THERMODYNAMICS

Why does the enthalpy of melting ice decrease
as the temperature decreases?

Temperature dependence of enthalpy

The enthalpy of freezing water is −6.00 kJ mol −1 at its normal freezing temperature
of 273.15 K. The value is negative because energy is liberated during freezing. But
the freezing temperature of water changes if the external pressure
The word ‘normal’ in is altered; so, for example, water freezes at the lower temperature
this context means ‘at of 253 K when a pressure of about 100 × p is applied. This high
O
a pressure of p ’. pressure is the same as that along the leading edge of an aeroplane
O
−1
wing. At this lower temperature, H (melt) = 5.2kJ mol .
The principal cause of H (melt) changing is the decreased temperature. The magni-
tude of an enthalpy depends on the temperature. For this reason, we need to cite the
temperature at which an enthalpy is determined. If the conditions are not cited, we
O
assume a temperature of 298 K and a pressure of p . We recognize these conditions
as s.t.p. Values of enthalpy are often written as H O for this reason.
r 298 K
The temperature dependence of the standard enthalpy is related by Kirchhoff’s law:

T 2
H O = H O + C p (T ) dT (3.19)
r T 2 r T 1
T 1
where C p is the molar heat capacity at constant pressure of the
Reminder: The ‘curly d’ substance in its standard state at a temperature of T . We define C p
symbols ∂ tells us the according to
bracketed term in the ∂H
equation is a ‘partial C p = ∂T (3.20)
differential’. p

The value of C p is itself a function of temperature (see p. 140),
C p (T) means C p as a which explains why we integrate C p (T ) rather than C p alone.
function of thermody- The Kirchhoff law is a direct consequence of the heat capacity at
namic temperature. constant pressure being the derivative of enthalpy with respect to
temperature. It is usually sufficient to assume that the heat capacity
C p is itself independent of temperature over the range of temperatures required, in
which case Equation (3.19) simplifies to

H O = H O + C p (T 2 − T 1 ) (3.21)
r T 2 r T 1

The experimental scientist should ensure the range of temperatures is slight if calcu-
lating with Equation (3.21).

Worked Example 3.9 The standard enthalpy of combustion H O for benzoic acid
c
◦
(I)is −3223.1kJ mol −1 at 20 C. What is H O ? The change in C p during the
c 298 K