Question

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

Answer 1

Answer:

For A: The expression for $K_{eq}$ is given below.

For B: The value of $K_{eq}$ at 25°C is 0.0185512

For C: The value of $K_{eq}$ at 65°C is 0.2887886

For D: The reaction is endothermic in nature.

Explanation:

• For A:

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 $K_{eq}$

For the given chemical reaction:

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

The expression of $K_{eq}$ for above equation without the concentration of liquid water is:

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

The expression is written above.

• For B:

We are given:

$[CoCl_4]^{2-}=0.0334612M$

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

$[Cl^-]=1.86616M$

Putting values in equation 1, we get:

$K_{eq}=\frac{0.0334612}{0.966539\times 1.86616}=0.0185512$

Hence, the value of $K_{eq}$ at 25°C is 0.0185512

• For C:

We are given:

$[CoCl_4]^{2-}=0.234625M$

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

$[Cl^-]=1.06150M$

Putting values in equation 1, we get:

$K_{eq}=\frac{0.234625}{0.765375\times 1.06150}=0.2887886$

Hence, the value of $K_{eq}$ at 65°C is 0.2887886

• For D:

For Endothermic reactions, $\Delta H>0$, which is positive

For Exothermic reactions, $\Delta H$, which is negative

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

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

where,

$K_{65^oC}$ = equilibrium constant at 65°C = 0.2887886

$K_{25^oC}$ = equilibrium constant at 25°C = 0.0185512

$\Delta H$ = Enthalpy change of the reaction = ?

R = Gas constant = 8.314 J/mol K

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

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

Putting values in above equation, we get:

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

As, the calculated value of $\Delta H>0$. Thus, the reaction is endothermic in nature.