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Question: For the reaction \[MX\left( s \right) \leftrightarrow {M^ + }\left( {aq} \right) + {X^ - }\left( {aq...

For the reaction MX(s)↔M+(aq)+X−(aq),ΔG∘=57kJMX\left( s \right) \leftrightarrow {M^ + }\left( {aq} \right) + {X^ - }\left( {aq} \right),\Delta {G^\circ } = 57kJ. Ksp{K_{sp}}
A. 10−5{10^{ - 5}}
B. 10−6{10^{ - 6}}
C. 10−10{10^{ - 10}}
D. 10−16{10^{ - 16}}

Explanation

Solution

ΔG∘\Delta {G^\circ } stands for Gibbs free energy, and Ksp{K_{sp}} stands for the solubility product of the given compound MX . There is a direct relationship between the solubility product and Gibbs free energy of a compound, which can be used to determine the value of Ksp{K_{sp}} for the compound MX . We must keep in mind that since no particular temperature condition is mentioned at which the reaction is taking place, the reaction is assumed to be occurring at the standard temperature of 25∘C{25^\circ }C itself.

Complete step by step answer:
The relationship between the solubility product and Gibbs free energy is given by the formula:
ΔG∘=−2.303×R×T×logKsp\Delta {G^\circ } = - 2.303 \times R \times T \times log{K_{sp}} …(i)(i)
Where R represents the ideal gas constant of reaction;
T represents the temperature of the reaction in Kelvin.

The given value of ΔG∘\Delta {G^\circ } for the dissociation reaction of MXMX is 57kJ57kJ.
All the values must be in the S.I. unit. So, we have to convert kJkJ into JJ (joules).
We know that, 1kJ = 1000J
⇒57kJ=57×1000J=57000J\Rightarrow 57kJ = 57 \times 1000J = 57000J
The value of RR in S.I. unit is the constant value of 8.314J/mol/K8.314J/mol/K.
The S.I. value of 25∘C{25^\circ }C in Kelvin =25+273K=298K = 25{{ }} + {{ }}273{{ }}K{{ }} = {{ }}298{{ }}K

Substituting all the values in the equation (i)(i) , we get;

57000  J/mol=−2.303×8.314J/mol/K×298K×logKsp57000\;J/mol = - 2.303 \times 8.314J/mol/K \times 298K \times log{K_{sp}}

⇒57000J/mol=−5705.84J/mol×logKsp \Rightarrow 57000J/mol = - 5705.84J/mol \times log{K_{sp}}

⇒logKsp=57000J/mol−5705.84J/mol \Rightarrow log{K_{sp}} = \dfrac{{57000J/mol}}{{ - 5705.84J/mol}}

⇒logKsp=−10 \Rightarrow log{K_{sp}} = - 10

⇒Ksp=10−10 \Rightarrow {K_{sp}} = {10^{ - 10}}
Therefore, the value of Ksp{K_{sp}} for the dissociation of MXMX using Gibbs free energy of 57kJ57kJ is 10−10{10^{ - 10}}.

So, the correct answer is Option C.

Additional Information:
Gibbs free energy is the energy available in a system that determines whether a reaction will take place in the system.
Solubility product is a measure or the extent to which a solute dissolves in a solvent after dissociation. Its value increases with the increase in solubility of a compound in a solvent.

Note: A reaction can take place only if Gibbs free energy for the reaction is negative. So, the negative sign in the equation must not be omitted.
To convert degree Celsius into Kelvin, we must add 273273 to it, as 1∘C=273K{1^\circ }C = 273K.
It must be kept in mind that the natural base of a logarithm is 1010. So, logKsp=−10log{K_{sp}} = - 10 means log10Ksp=−10lo{g_{10}}{K_{sp}} = - 10.