[Co(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CoCl4]2-(aq) + 6H2O(l)

Concentrations at 25 degrees C

H+ 9.84368e-8

OH- 9.84368e-8

Co (H2O)6^2 0.966539

Cl- 1.86616

CoCl4^2- 0.0334612

Concentrations at 65 degree C

H+ 3.73844e-7

OH- 3.73844e-7

Co (H2O)6^2 0.765375

Cl- 1.06150

CoCl4^2- 0.234625

A. Write a K expression for this reaction (note that that liquid water shouldn’t be included in the K expression).

B. Use the K expression and equilibrium concentrations on the left to determine the K value at 25 deg C. Please show all work for full credits.

C. What is the equilibrium constant K’ 65 Degrees C?

D. Compare the K’ constants; are the value difference agree or against with endo or exothermic determination?

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{eq}[/tex]

For the given chemical reaction:

[tex][Co(H_2O)_6]^{2+}(aq.)+4Cl^-(aq.)\rightleftharpoons [CoCl_4]^{2-}(aq.)+6H_2O(l)[/tex]

The expression of [tex]K_{eq}[/tex] for above equation without the concentration of liquid water is:

[tex]K_{eq}=\frac{[CoCl_4]^{2-}}{[Co(H_2O)_6]^{2+}[Cl^-]^4}[/tex] ......(1)

The expression is written above.

For B:

We are given:

[tex][CoCl_4]^{2-}=0.0334612M[/tex]

[tex][Co(H_2O)_6]^{2+}=0.966539M[/tex]

[tex][Cl^-]=1.86616M[/tex]

Putting values in equation 1, we get:

[tex]K_{eq}=\frac{0.0334612}{0.966539\times 1.86616}=0.0185512[/tex]

Hence, the value of [tex]K_{eq}[/tex] at 25°C is 0.0185512

For C:

We are given:

[tex][CoCl_4]^{2-}=0.234625M[/tex]

[tex][Co(H_2O)_6]^{2+}=0.765375M[/tex]

[tex][Cl^-]=1.06150M[/tex]

Putting values in equation 1, we get:

[tex]K_{eq}=\frac{0.234625}{0.765375\times 1.06150}=0.2887886[/tex]

Hence, the value of [tex]K_{eq}[/tex] at 65°C is 0.2887886

For D:

For Endothermic reactions, [tex]\Delta H>0[/tex], which is positive

For Exothermic reactions, [tex]\Delta H<0[/tex], which is negative

To calculate [tex]\Delta H[/tex] of the reaction, we use Van't Hoff's equation, which is:

[tex]\ln(\frac{K_{65^oC}}{K_{25^oC}})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_{65^oC}[/tex] = equilibrium constant at 65°C = 0.2887886

[tex]K_{25^oC}[/tex] = equilibrium constant at 25°C = 0.0185512

[tex]\Delta H[/tex] = Enthalpy change of the reaction = ?

R = Gas constant = 8.314 J/mol K

[tex]T_1[/tex] = initial temperature = [tex]25^oC=[25+2730]K=298K[/tex]

[tex]T_2[/tex] = final temperature = [tex]65^oC=[65+2730]K=338K[/tex]

Putting values in above equation, we get:

[tex]\ln(\frac{0.2887886}{0.0185512})=\frac{\Delta H}{8.314J/mol.K}[\frac{1}{298}-\frac{1}{338}]\\\\\Delta H=57471.26J/mol[/tex]

As, the calculated value of [tex]\Delta H>0[/tex]. Thus, the reaction is endothermic in nature.