Determine the equilibrium constant for the following reaction at 527 K. 2Hg(g)+O2(g)-->2HgO(g) Delta H^o= -304.2 kJ...

Question

Question

Determine the equilibrium constant for the following reaction at 527 K.

2Hg(g)+O2(g)-->2HgO(g)

Delta H^o= -304.2 kJ and Delta S^o= -414.2 J/K

Answer 1

Answer #1

First we will calculate the change in free energy using following equation

Delta G^o = Delta H^o - T Delta S^o

Delta G^o = - 304200 J - { 527 K * (-414 J/K) } = -304200 J - { - 218178 J } = - 86022 J.

Now we can apply the relation between standard free energy change (Delta G^o) and equilibrium constant K

Delta G^o = - 2.303 RT log K

- 86022 J = - 2.303 * 8.314 * 527 * log K

log K = 86022 / 10090.5 = 8.525

Now taking antilog of log K

Eqilibrium constant K = 3.35 * 10⁸

Answer 2

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