Question

# Derive the relation between $$K_p$$ and $$K_c$$ for a general chemical equilibrium reaction.

Solution

## Let the gaseous reaction is a state of equilibrium is-$$a{A}_{\left( g \right)} + b{B}_{\left( g \right)} \rightleftharpoons c{C}_{\left( g \right)} + d{D}_{\left( g \right)}$$Let $${p}_{A}, {p}_{B}, {p}_{C}$$ and $${p}_{D}$$ be the partial pressure of $$A, B, C$$ and $$D$$ repectively.Therefore,$${K}_{c} = \cfrac{{\left[ C \right]}^{c} \; {\left[ D \right]}^{d}}{{\left[ A \right]}^{a} {\left[ B \right]}^{b}} ..... \left( 1 \right)$$$${K}_{p} = \cfrac{{{p}_{C}}^{c} \; {{p}_{D}}^{d}}{{{p}_{A}}^{a} \; {{p}_{B}}^{b}} ..... \left( 2 \right)$$For an ideal gas-$$PV = nRT$$$$\Rightarrow P = \cfrac{n}{V} RT = CRT$$Whereas $$C$$ is the concentration.Therefore,$${p}_{A} = \left[ A \right] RT$$$${p}_{B} = \left[ B \right] RT$$$${p}_{C} = \left[ C \right] RT$$$${p}_{D} = \left[ D \right] RT$$Substituting the values in equation $$\left( 2 \right)$$, we have$${K}_{p} = \cfrac{{\left[ C \right]}^{c} {\left( RT \right)}^{c} \; {\left[ D \right]}^{d} {\left( RT \right)}^{d}}{{\left[ A \right]}^{a}{\left( RT \right)}^{a} {\left[ B \right]}^{b} {\left( RT \right)}^{b}}$$$$\Rightarrow {K}_{p} = \cfrac{{\left[ C \right]}^{c} \; {\left[ D \right]}^{d}}{{\left[ A \right]}^{a} {\left[ B \right]}^{b}} {\left( RT \right)}^{\left[ \left( c + d \right) - \left( a + b \right) \right]}$$$$\Rightarrow {K}_{p} = {K}_{c} {\left( RT \right)}^{\Delta{{n}_{g}}} \quad \left( \text{From } \left( 1 \right) \right]$$Here,$$\Delta{{n}_{g}} =$$ Total no. of moles of gaseous product $$-$$ Total no. of moles of gaseous reactantHence the relation between $${K}_{p}$$ and $${K}_{c}$$ is-$${K}_{p} = {K}_{c} {\left( RT \right)}^{\Delta{{n}_{g}}}$$Chemistry

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