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Dahasolnce [82]
4 years ago
12

What occurs in an ionic bond ? and why ?

Chemistry
1 answer:
BabaBlast [244]4 years ago
6 0
A metal and nonmetal are bonded together by exchanging electrons. I'm not sure what you mean by "why".

Hope I helped! :)
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Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.
VladimirAG [237]

<u>Answer:</u>

<u>For 1:</u> The standard Gibbs free energy change of the reaction is 10.60 kJ/mol

<u>For 2:</u> The equilibrium constant for the given reaction at 298 K is 1.386\times 10^{-2}

<u>For 3:</u> The equilibrium pressure of oxygen gas is 0.0577 atm

<u>Explanation:</u>

  • <u>For 1:</u>

The equation used to calculate standard Gibbs free energy change of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]

For the given chemical reaction:

M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)

The equation for the standard Gibbs free energy change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})]

We are given:

\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol

Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol

  • <u>For 2:</u>

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol    (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 298 K

K_{eq} = equilibrium constant = ?

Putting values in above equation, we get:

10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}

Hence, the equilibrium constant for the given reaction at 298 K is 1.386\times 10^{-2}

  • <u>For 3:</u>

The expression of K_{eq} for above equation follows:

K_{eq}=p_{O_2}^{3/2}

The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.

Putting values in above expression, we get:

1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm

Hence, the equilibrium pressure of oxygen gas is 0.0577 atm

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