Q20.70 P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate K at $\mathbf{\text{298 K}}$  for each reaction:

(a)  ${\mathbf{\text{2H}}}_{\mathbf{\text{2}}}{\mathbf{\text{ S(g)+3O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\rightleftharpoons }{\mathbf{\text{2H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O(g)+2SO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

(b)  ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{SO}}}_{\mathbf{\text{4}}}\mathbf{\text{(l)}}\mathbf{\rightleftharpoons }{\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O(l)+SO}}}_{\mathbf{\text{3}}}\mathbf{\text{(g)}}$

(c)  $\mathbf{\text{HCN(aq)+NaOH(aq)}}\mathbf{\rightleftharpoons }{\mathbf{\text{NaCN(aq)+H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(l)}}$

Step-by-Step Solution

Answer

(a) The value is ${\text{6.6×10}}^{\text{173}}$ .

(b) The value is  ${\text{4.5×10}}^{\text{-15}}$ .

(c) The value is  ${\text{3.5×10}}^{\text{4}}$.

Step 1: Definition of Concept

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

Step 2: Calculate K

(a)

Consider the reaction given below,

 ${\text{2H}}_{\text{2}}{\text{ S(g)+3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2H}}_{\text{2}}{\text{O(g)+2SO}}_{\text{2}}\text{(g)}$

To solve for K,

 we can use the equation below.

 ${\text{ΔG}}^{\text{°}}\text{=-RTlnK}$

First, we have to solve for  using the Gibb's free energy constants found in Appendix 

B.

 $\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{=}\sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{( product )-}\sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{( reactant )}\\ \text{ =}\left[{\text{nΔG}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(g)}\right){\text{+nΔG}}_{\text{f}}^{\text{°}}\left({\text{SO}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{nΔG}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{S(g)}\right){\text{+nΔG}}_{\text{f}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\right]\\ \text{ =([2(-228.6)+2(-300.2)]-[2(-33)+3(0)])kJ/mol}\\ {\text{ΔG}}^{\text{°}}\text{=-991.6 kJ/mol}\end{array}$

Then solve for K.

 $\begin{array}{c}{\text{ ΔG}}^{\text{°}}\text{=-RTlnK}\\ {\text{ K=e}}^{\frac{{\text{Δc}}^{\text{°}}}{\text{KI}}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{-991.6 kJ/mol×}\frac{\text{1000 J}}{\text{1 kJ}}}{\text{-8.314 J/mol×K×298 K}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=400.23}\\ {\text{ K=e}}^{\text{400.23}}\\ {\text{ K=6.6×10}}^{\text{173}}\end{array}$

Therefore, the required value is  ${\text{6.6×10}}^{\text{173}}$.

Step 3: Calculate K

(b)

Consider the reaction given below,

 ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(l)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{O(l)+SO}}_{\text{3}}\text{(g)}$

To solve for K, 

we can use the equation below.

 ${\text{ΔG}}^{\text{°}}\text{=-RTlnK}$

First, we have to solve for ${\text{ΔG}}^{\text{°}}$ using the Gibb's free energy constants found in Appendix

B.

$\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{=}\sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{( product )-}\sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{( reactant )}\\ \text{ =}\left[{\text{nΔG}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(l)}\right){\text{+nΔG}}_{\text{f}}^{\text{°}}\left({\text{SO}}_{\text{3}}\text{(g)}\right)\right]\text{-}\left[{\text{nΔG}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(l)}\right)\right]\\ \text{ =([1(-237.192)+1(-371)]-[1(-690.059)])kJ/mol}\\ {\text{ΔG}}^{\text{°}}\text{=81.867 kJ/mol}\end{array}$

Then solve for K.

 $\begin{array}{c}{\text{ ΔG}}^{\text{°}}\text{=-RTlnK}\\ {\text{ K=e}}^{\frac{{\text{ΔC}}^{\text{°}}}{\text{KI}}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{81.867 kJ/mol×}\frac{\text{1000 J}}{\text{1kJ}}}{\text{-8.314 J/mol×K×298K}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=-33.04}\\ {\text{ K=e}}^{\text{-33.04}}\\ {\text{ K=4.5×10}}^{\text{-15}}\end{array}$

Therefore, the required value is ${\text{4.5×10}}^{\text{-15}}$ .

Step 4: Calculate K

(c)

Consider the reaction given below,

 $\text{HCN(aq)+NaOH(aq)}\rightleftharpoons {\text{NaCN(aq)+H}}_{\text{2}}\text{O(l)}$

 ${\text{HCN(aq)+Na}}^{\text{+}}{\text{(aq)+OH}}^{\text{-}}\text{(aq)}\rightleftharpoons {\text{Na}}^{\text{+}}{\text{(aq)+CN}}^{\text{-}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}$

${\text{Na}}^{\text{+}}$ is just a spectator ion.

 ${\text{HCN(aq)+OH}}^{\text{-}}\text{(aq)}\rightleftharpoons {\text{+CN}}^{\text{-}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}$

To solve for K, we can use the equation below.

 ${\text{ΔG}}^{\text{°}}\text{=-RTlnK}$

First, we have to solve for  using the Gibb's free energy constants found in Appendix 

B.

$\begin{array}{c}{\text{ΔG}}^{\text{°}}\text{=}\sum {\text{n}}_{\text{p}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{( product )-}\sum {\text{n}}_{\text{r}}{\text{ΔG}}_{\text{f}}^{\text{°}}\text{( reactant )}\\ \text{ =}\left[{\text{nΔG}}_{\text{f}}^{\text{°}}\left({\text{CN}}^{\text{-}}\text{(aq)}\right){\text{+nΔG}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{O(l)}\right)\right]\text{-}\left[{\text{nΔG}}_{\text{f}}^{\text{°}}{\text{(HCN(aq))+nΔG}}_{\text{f}}^{\text{°}}\left({\text{OH}}^{\text{-}}\text{(aq)}\right)\right]\\ \text{ =([1(166)+1(-237.192)]-[1(112)+1(-157.30)])kJ/mol}\\ {\text{ΔG}}^{\text{°}}\text{=-25.892 kJ/mol}\end{array}$

Then solve for K.

 $\begin{array}{c}{\text{ ΔG}}^{\text{°}}\text{=-RTlnK}\\ {\text{ K=e}}^{\frac{}{\text{G}}\text{×}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=}\frac{\text{-25.892 kJ/mol×}\frac{\text{1000 J}}{\text{1 kJ}}}{\text{-8.314 J/mol×K×298 K}}\\ \frac{{\text{ΔG}}^{\text{°}}}{\text{-RT}}\text{=10.45}\\ {\text{ K=e}}^{\text{10.45}}\\ {\text{ K=3.5×10}}^{\text{4}}\end{array}$

Therefore, the required value is  ${\text{3.5×10}}^{\text{4}}$.