Question

1) For iodine gas [ I2(g) ], delta H= 62.4 kJ/mol and S = 260.7 J/mol-K. Calculate the equilibrium partial pressures of I2(g), H2(g), and HI(g) for the system

2HI(g) ⇌ H2(g) + I2(g)
at 500oC if the initial partial pressures are all 0.200 atm.

Answer 1

using the thermodynamic equation

deltaG = deltaH - T deltaS

we get

deltaG = 62.4 kJ/mol - ( 25 + 273 ) K * 260.7 J/mol-K

= 62400 J/mol - J/mol

= - 15288.6 J/mol

and

deltaG = - 2.303 RT logKeq gives

logKeq = - 15288.6 J/mol / ( - 2.303 * 8.314 * 773 )

= 1.033

Keq = 10^1.033

= 10.79

therefore for the reaction

2HI(g) ⇌ H2(g) + I2(g)

0.2 atm - 2x 0.2 + x 0.2 + x

Keq = ( 0.2 + x )^2 / ( 0.2 - 2x ) ^2

10.79 = ( 0.2 + x )^2 / ( 0.2 - 2 x )^2   

( 0.2 + x ) / ( 0.2 - 2 x ) = 3.28

0.2 + x = 0.66 - 6.56 x

7.56 x = 0.66 - 0.2

x = 0.06 atm

therefore

[HI] = 0.2 - 2 * 0.06 = 0.08 atm

[I2] = 0.2 + 0.06 = 0.26 atm

[H2] = 0.2 + 0.06 = 0.26 atm