1. Xtra Edge February 2012 - Career Point

1. It is possible to supercool water without freezing. 18
g of water are supercooled to 263.15 K(–10ºC) in a
thermostat held at this temperature, and then
crystallization takes place.
Calculate ∆rG for this process. Given:
Cp(H2O,1) = 75.312 J K –1 mol –1
Cp (H2O,s) = 36.400 J K –1 mol –1
∆fusH (at 0ºC) = 6.008 kJ mol –1
Sol. The process of crystallization at 0ºC and at 101.325
kPa pressure is an equilibrium process, for which
∆G = 0. The crystallization of supercooled water is a
spontaneous phase transformation, for which ∆G
must be less than zero. Its value for this process can
be calculated as shown below.
The given process
H2O(1, – 10ºC) → H2O(s, –10ºC)
is replaced by the following reversible steps.
(a) H2O(1, – 10ºC) → H2O(1, 0ºC) ...(1)
273.
15K
∆rH1 = ∫ Cp , m ( 1)
dT
263.
15K
= (75.312 J K –1 mol –1 ) (10 K)
= 753.12 J mol –1
273.
15K
∆rS1 = ∫
263.
15K
C
p
, m
R
( 1)
dT
= (75.312 J K –1 mol –1 ⎛ 273.
15K
⎞
) × ln
⎜
⎟
⎝ 263.
15K
⎠
= 2.809 J K –1 mol –1
(b) H2O(1, 0ºC) → H2O(s, 0ºC) ...(2)
∆rH2 = – 6.008 kJ mol –1
– 1
( 6008 J mol )
∆rS2 = –
= – 21.995 J K
( 273.
15 K)
–1 mol –1
(c) H2O(s, 0ºC) → H2O(s, –10ºC) ...(3)
263.
15K
∆rH3 = ∫ Cp , m ( s)
dT
273.
15K
= (36.400 J K –1 mol –1 )(–10 K)
= – 364.0 J mol –1
263.
15K
∆rS3 = ∫
273.
15K
C
p,
m
T
( s)
dT
UNDERSTANDING
= (36.400 J K –1 mol –1 ⎛ 263.
15K
⎞
) ×ln
⎜
⎟
⎝ 273.
15K
⎠
= – 1.358 J K –1 mol –1
The overall process is obtained by adding Eqs. (1),
(2) and (3), i.e.
H2O(1, –10ºC) → H2O(s, –10ºC)
The total changes in ∆rH and ∆rS are given by
∆rH = ∆rH1 + ∆rH2 + ∆rH3
=(753.12 – 6008 – 364.0) J mol –1
= – 5618.88 J mol –1
∆rS = ∆rS1 + ∆rS2 + ∆rS3
= (2.809 – 21.995 – 1.358) J K –1 mol –1
= – 20.544 J K –1 mol –1
Now ∆rG of this process is given by
∆rG = ∆rH – T∆rS
= – 5618.88 J mol –1 – (263.15 K)( –20.544 J K –1 mol –1 )
= – 212.726 J mol –1
2. From the standard potentials shown in the following
diagram, calculate the potentials
º
E 1 and
XtraEdge for IIT-JEE 33 FEBRUARY 2012
E1º
º
E 2 .
BrO3 – 0.54 V –
BrO
0.45 V 1 1.07 V
Br2
2
–
Br
0.17 V
E2º
Sol. The reaction corresponding to the potential Eº1 is
BrO3 – + 3H2O + 5e – 1
= Br2 + 6OH
2
– ...(1)
This reaction can be obtained by adding the
following two reduction reactions:
BrO3 – + 2H2O + 4e – = BrO – + 4OH – ...(2)
BrO – + H2O + e – 1
= Br2 + 2OH
2
–
...(3)
Hence the free energy change of reaction (1) will be
º
reaction(
1)
Physical Chemistry
º
reaction(
2)
∆ G = ∆ G +
∆ G
º
reaction(
3)
Replacing ∆Gºs in terms of potentials, we get
– 5FE1º = – 4F(0.54 V) – 1F (0.45 V)
= (–2.61 V) F