Question

Find the value of equilibrium constant in temperature 300 degree celcius of following equation by using vant hoff equation.

H2(g) +I2(s) --------- 2HI(g) Del H = 52.96

Answer 1

Given Thermochemical equation,

H₂(g) +I₂(s) --------- 2HI(g)

H^o = 52.96 kJ/mole = 52960 J/mole.

Let us calculate S^o using the thermodynamic table of Standard Molar entropy values.

S^o = Sum of entropies of Products - Sum of entropies of reactants

  S^o = 2 x S_f^o (HI)g - [S_f^o(H₂)g - S_f^o(I₂)s]

  S^o = 2 x (206.6) - [130.7 + 116.1]

  S^o = 166.4 J/mole*K

Then, Assuming that  S^oS^o remains constant over a range of temperature we have

G= H^o- T S^o.

T = 300 K

G= 52960 - 300 x 166.4

G^o = 3040 J

Then we know that the equilibrium constant K and  Gare related as,

G^o = -RT lnK

K = e^{-( G/RT)}.

K = e^{-(3040 / 8.314*300)}

K = e^-(1.22)

K = 0.3

Equilibrium constant for given equation is 0.3.

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