Page 60 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

44 Chapter 4

97.8 f 6(-97.8)2 - (4 x 44.9 x 48.9)
Hence: x =
2 x 44.9
= (97.8 f 27.971)/89.9 This generates two values of x: 0.777
and 1.399. Since x must be less than 1, the correct answer is
0.777.
9. Determine the equilibrium concentrations:
[H2] = 1 - x = 1 - 0.777 = 0.223 M; [I21 = 0.223 M; [HI] =
2x = 2 x 0.777 = 1.554 M.
10. One final note. In this question, the assumption, x must be << 1
was invalid, and a quadratic equation was used. However, on
close examination of (t), if the square root is taken on both
sides of the equation, a linear equation is obtained, from which
x can be evaluated. The expressions generated should always be
closely examined for potential short-cuts in such questions!
11. Kp = P(HI>2/(P(H21P(12)> = 48.9.

RELATIONSHIP BETWEEN AG AND Kp

In Chapter 3, AGO, the change in standard Gibbs free energy, was
related to both the change in the standard enthalpy and the change in
the standard entropy:
i.e. AGO = AH" - TAS"
AG is also related to Kp from the expression:
r

AG = AGO + RTlnK,

At equilibrium, AG = 0
i.e. AGO + RTln Kp = 0
RTlnK, = -AGO +- lnKp = (-AGo)/(RT).
Hence, if AHo and AS" can be calculated for a reaction, AGO can be
determined from the equation above, and ultimately a value for Kp,
the equilibrium constant, can be obtained.

LE CH~TELIER'S PRINCIPLE
Le Chiitelier studied the influence of temperature, pressure and
concentration on systems at equilibrium.