Question

Calculate the temperature at which the reaction given below is at equilibrium.

Ag2o(s)⇔ 2Ag(s) +1/2O2(g)

ΔH°= 30.5 kJmol^-1
ΔS°=0.006 kJmol^-1

Answer 1

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Answer 2

Answer:

The temperature at which the reaction given below is at equilibrium is 5,083.33 K.

Explanation:

$Ag2O(s)\rightleftharpoons 2Ag(s) +\frac{1}{2}O_2(g)$

The Gibbs free energy expression is given as:

$\Delta G=\Delta H-T\Delta S$

At equilibrium the $\Delta G$ is equal to zero.

$0 J/mol=30.5kJ /mol-T\times 0.006 kJ/mol$

$T=\frac{\Delta H}{\Delta S}=\frac{30.5 kj/mol}{0.006 kJ/mol}=5,083.33 K$

The temperature at which the reaction given below is at equilibrium is 5,083.33 K.