Question

Convert the values of K_c to values of K_p or the values of K_p to values of K_c .

\((a)\,{N_2}\left( g \right) + 3{H_2}\left( g \right)\rightleftharpoons 2N{H_3}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_C} = 0.50\,\,at\,\,400^\circ C\)

\((b){{\rm{H}}_2}(g) + {{\rm{I}}_2}(g)\rightleftharpoons 2HI(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_c} = 50.2\,at\,{448^\circ }{\rm{C}}\)

\((c)N{a_2}{\rm{S}}{{\rm{O}}_4} \cdot 10{{\rm{H}}_2}O(s)\rightleftharpoons N{a_2}{\rm{S}}{{\rm{O}}_4}(s) + 10{{\rm{H}}_2}O(g){K_P} = 4.08 \times {10^{ - 25}}at\,{25^\circ }{\rm{C}}\)

\((d){{\rm{H}}_2}{\rm{O}}(l)\rightleftharpoons {{\rm{H}}_2}{\rm{O}}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_P} = 0.122\,at\,{50^\circ }{\rm{C}}\)

Short answer

\(\begin{aligned}{}a)\,{K_p} &= 1.6 \times {10^{ - 8}}\\b)\,{K_p} &= 50.2\\c)\,{K_c} &= 4.7 \times {10^{ - 59}}\\d)\,{K_c} &= 542.9\end{aligned}\)

Step-by-step solution

 Step 1: To calculate\({\rm{Kp}}\) from \({\rm{Kc}}\)

To calculate K_p from K_cor vice versa, we will use the following equation:

\({K_p} = {K_c} \cdot {(RT)^{\Delta n}}\)

Where\(R\)is gas constant\(\left( {8.314J{K^{ - 1}}mo{l^{ - 1}}} \right)\)and\(T\)is thermodynamic temperature.\(\Delta n\)can be defined for this hypothetical reaction:

\(aA(g) + bB(g) \to cC(g) + dD(g)\)

\(\Delta n = c + d - (a + b)\)

\((a,b,c,d\)are stoichiometry coefficients but only for the gas reactants or products).

Determine \({\rm{Kp}}\) for part (a)  

a) Here \(\Delta n\) is:

\(\Delta n = 2 - 1 - 3\)

\(\Delta n = - 2\)

Thermodynamic temperature is calculated like this:

\(T = t + 273.15\)

\(T = (400 + 273.15)K\)

\(T = 673.15K\)

Now we can calculate\({K_p}\):

\({K_p} = 0.5 \times {\left( {673.15\;{\rm{K}} \times 8.314J{K^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}} \right)^{ - 2}}\)

\({K_p} = 0.5 \times 3.2 \times {10^{ - 8}}\)

\({K_p} = 1.6 \times {10^{ - 8}}\)

Determine \({\rm{Kp}}\) for part (b)

b) Here \(\Delta n\) is:

\(\Delta n = 2 - 1 - 1\)

\(\Delta n = 0\)

Thermodynamic temperature is:

\(T = t + 273.15\)

\(T = (448 + 273.15)K\)

\(T = 721.15K\)

Now we can calculate\({K_p}\):

\({K_p} = 50.2 \times {\left( {721.15K \times 8.314{J^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}} \right)^0}\)

\({K_p} = 50.2 \times 1\)

\({K_p} = 50.2\)

Determine \({{\rm{K}}_{\rm{C}}}\) for part (c)

c)

\(\Delta n = 10 + 0 - 0\)

(Remember only\(\Delta n\)is calculated only for the gas reactants and products)

\(\Delta n = 10\)

Thermodynamic temperature is:

\(T = t + 273.15\)\(\)

\(T = 298.15K\)

Now we can calculate\({K_c}\):

\({K_p} = {K_c} \cdot {(RT)^{\Delta n}}\)

\({K_c} = \frac{{{K_p}}}{{{{(RT)}^{\Delta n}}}}\)

\({K_c} = \frac{{4.08 \times {{10}^{ - 25}}}}{{{{\left( {298.15K \times 8.314{J^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}} \right)}^{10}}}}\)

\({K_c} = \frac{{4.08 \times {{10}^{ - 25}}}}{{8.76 \times {{10}^{33}}}}\)

\({K_c} = 4.7 \times {10^{ - 59}}\)

Determine \({{\rm{K}}_{\rm{C}}}\) for part (d)

d)

\(\Delta n = 1 - 0\)

(Remember only\(\Delta n\)is calculated only for the gas reactants and products)

\(\Delta n = 1\)

Thermodynamic temperature is:

\(T = t + 273.15\)

\(T = (60 + 273.15)K\)

\(T = 333.15K\)

Now we can calculate\({K_c}\):

\({K_p} = {K_c} \cdot {(RT)^{\Delta n}}\)

\({K_c} = \frac{{{K_p}}}{{{{(RT)}^{\Delta n}}}}\)

\({K_c} = \frac{{0.196}}{{{{\left( {333.15K \times 8.314J{K^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}} \right)}^1}}}\)

\({K_c} = \frac{{0.196}}{{{{2769.8}^1}}}\)

\({K_c} = 542.9\)