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zhannawk [14.2K]
3 years ago
8

Which is the reason that ionic compounds are brittle?

Chemistry
2 answers:
Vaselesa [24]3 years ago
8 0

Answer:

<u><em>The answer is</em></u>: <u>B. A strong repulsive force is created when the cations suddenly line up.</u>

Explanation:

<em>Solid ionic compounds are fragile. </em>This behavior is explained as a result of the repulsions that appear when one layer of ions moves with respect to the other<em> so that ions of the same charge are stopped.</em>

<u><em>The answer is</em></u>: <u>B. A strong repulsive force is created when the cations suddenly line up.</u>

lisabon 2012 [21]3 years ago
7 0
Ionic bonds are the weakest bonds. 
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In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a
igor_vitrenko [27]

<u>Answer:</u> The equilibrium constant for this reaction is 1.068\times 10^{6}

<u>Explanation:</u>

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

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

For the given chemical reaction:

Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)

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

\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]

We are given:

\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol

To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:

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

where,

\Delta G^o = Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol  (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 58^oC=[273+58]K=331K

K_{eq} = equilibrium constant at 58°C = ?

Putting values in above equation, we get:

-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}

Hence, the equilibrium constant for this reaction is 1.068\times 10^{6}

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