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Q.

What is the equation for the equilibrium constant (KC) for the following reaction ? \large \frac{1}{2}A(g)\, + \,\frac{1}{3}B(g)\,\, \rightleftharpoons \,\frac{2}{3}C(g)

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a

{K_c}\, = \,\frac{{{{[C]}^{\frac{2}{3}}}}}{{{{[A]}^{\frac{1}{2}}}{{[B]}^{\frac{1}{3}}}}}

 

b

{K_c}\, = \,\frac{{{{[A]}^{\frac{1}{2}}}{{[B]}^{\frac{1}{3}}}}}{{{{[C]}^{\frac{3}{2}}}}}

 

c

{K_c}\, = \,\frac{{{{[C]}^{\frac{3}{2}}}}}{{{{[A]}^2}{{[B]}^3}}}

 

d

{K_c}\, = \,\frac{{{{[C]}^{\frac{2}{3}}}}}{{{{[A]}^{\frac{1}{2}}} + {{[B]}^{\frac{1}{3}}}}}

 

answer is C.

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Detailed Solution

{K_C} = \frac{{\left[ {{\text{Product}}} \right]}}{{\left[ {{\text{Reactants}}} \right]}};\;aA(g) + bB(g) \rightleftharpoons mC(g) + nD\left( g \right)

 

{K_C} = \frac{{{{\left[ C \right]}^m}{{\left[ D \right]}^n}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}

 

 

\therefore for\;the\;equation\;\frac{1}{2}\;A(g)+\frac{1}{3}B(g) \rightleftharpoons \frac{2}{3}C(g)
{K_C} = \frac{{{{\left[ C \right]}^{\frac{2}{3}}}}}{{{{\left[ A \right]}^{\frac{1}{2}}}{{\left[ B \right]}^{\frac{1}{3}}}}}
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