Question

Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M3O4(s)↽−−⇀ 3M(s)+2O2(g) What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? ΔG∘rxn= kJ/mol What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K? K= What is the equilibrium pressure of O2(g) over M(s) at 298 K?

Answer 1

Answer:

The equilibrium constant is  $K =0.02867$

The equilibrium pressure for oxygen gas is  $P_{O_2} = 0.09367\ atm$

Explanation:

  From the question we are told that

       The equation of the chemical reaction is

                    $M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

   The Gibbs free energy for$M_3 O_4_{(s)}$ is  $\Delta G^o_{1} = -8.80 \ kJ/mol$

  The Gibbs free energy for$M{(s)}$ is  $\Delta G^o_{2} = 0 \ kJ/mol$

    The Gibbs free energy for$O_2{(s)}$ is  $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

         $\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

         $\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

         $\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

        $\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

           $\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of  $R = 8.314 J/mol \cdot K$

             T is the temperature with a given value  of  $T = 298 K$

             K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction  is mathematically represented as

                          $K_p =[ P_{O_2}]^{\frac{3}{2} }$

Where   $[ P_{O_2}]$ is the equilibrium pressure of oxygen

         The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the  second equation of the Gibbs free energy of the reaction

 

           $K = e^{- \frac{\Delta G^o_{re}}{RT} }$

Substituting values

           $K= e^{\frac{8800}{8.314 * 298} }$

            $K =0.02867$

Now substituting this into the equation above to obtain the equilibrium of oxygen

           $0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}$

multiplying through by $1 ^{\frac{2}{3} }$

        $P_{O_2} = [0.02867]^{\frac{2}{3} }$

        $P_{O_2} = 0.09367\ atm$