If you find that [CoCl42-(aq)] = 6.447 M, [Co(H2O)62+(aq)] = 0.113 M, and [Cl-(aq)] = 0.854 M at 6 ∘C, what is value of ΔG at 6 ∘C, in kJ/mol?


Given is the [CoCl42-(aq)] = 6.447M

[Co (H2O)2+(aq)] =0.113M

[Cl]=0.854M

Temperature = 6oC

We found out the ΔG at 6oC temperature in kJmol-1

Explanation:

We use the Gibbs free energy concept at equilibrium because the given data is for the reaction at equilibrium.

Step 2 of 2

[CoCl42-(aq)] = 6.447M

[Co (H2O)2+(aq)] =0.113M

[Cl]=0.854M

Temperature = 6 + 273= 279 K

As SI unit of temperature is K.

We found out the ΔG at 6oC=279K temperature in kJmol-1

Chemical reaction is given as:

[Co (H2O)2+(aq)] + 4[Cl(aq)] ↽−−⇀[CoCl42-(aq)] + 6H2O(l)

Keq=[CoCl42-(aq)] [H2O]6 ÷ [Co (H2O)2+(aq)] [Cl]4

Note – We ignore the concentration value of water [H2O] as it is present in excess quantity.

Keq=6.447÷0.113×(0.854)4 = 6.447÷0.06010491=107.262

By using equation of Gibbs free energy, ΔG=ΔGO+ RTlnQ

At equilibrium condition, Q=Keq and ΔG=0

So, equation ΔG=ΔGO+RTlnQ

Becomes 0=ΔGO+RTlnKeq

& ΔGO=-RTlnKeq

& ΔGO=-RT.2.303logKeq

& ΔGO=-2.303(8.314Jmol-1K-1) (279K) log (107.262)

& ΔGO=-5342.052618×2.030446 Jmol-1

& ΔGO=-10846.7493 Jmol-1

& ΔGO=-10.8467493 kJmol-1

Explanation:

Hence the ΔG is zero but ΔG 0 is not zero and has value of -10.847 kJmol-1

Final solution

Hence the ΔG is zero but ΔG 0 is not zero and has value of -10.847 kJmol-1

As the ΔG is zero but ΔG 0 <0, Keq>1; So, this reaction favors products formation.

 

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