Question Number 7884 by uchechukwu okorie favour last updated on 23/Sep/16
$${pls}\:{help}\:{me}\:{solve}\:{this}\:{chemistry} \\ $$$${question}: \\ $$$${Considering}\:{a}\:{general}\:{equilibrium} \\ $$$${equation}: \\ $$$${mA}_{\left({g}\right)} +{nB}_{\left({g}\right)} \leftrightharpoons{xC}_{\left({g}\right)\:\:} +{yD}_{\left({g}\right)} \\ $$$${where}\:{m},{n},{x}\:{and}\:{y}\:{are}\:{the}\: \\ $$$${coefficient}\:{in}\:{the}\:{balanced}\:{eqn}. \\ $$$${show}\:{that}:\:{K}_{{p}} ={K}_{{c}} \left({RT}\right)^{\left({x}+{y}\right)−\left({m}+{n}\right)} \\ $$$${where};\:{K}_{{c}} \:{is}\:{equilibrium}\:{constant} \\ $$$${and}\:{K}_{{p}} \:{is}\:{equilibrium}\:{constant} \\ $$$${when}\:{P}\:=\left[{gas}\:{conc}\right]{RT} \\ $$$${NB}:\:{P}={partial}\:{pressure} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{T}={temperature}\:{in}\:{kevin} \\ $$$$\:{and}\:{R}={gas}\:{constant} \\ $$$$ \\ $$
Answered by sandy_suhendra last updated on 23/Sep/16
$${from}\:\:{PV}={nRT}\:\:\Rightarrow\:{P}=\frac{{n}}{{V}}{RT} \\ $$$${where}\:\frac{{n}}{{V}}=\left[{gas}\right] \\ $$$${so}\:{P}_{{A}} \:=\:\left[{A}\right]{RT} \\ $$$$\:\:\:\:\:\:{P}_{{B}} \:=\:\left[{B}\right]{RT} \\ $$$$\:\:\:\:\:\:{P}_{{C}} \:=\:\left[{C}\right]{RT} \\ $$$$\:\:\:\:\:\:{P}_{{D}} \:=\:\left[{D}\right]{RT} \\ $$$$ \\ $$$${K}_{{P}} \:=\:\frac{\left[{P}_{{C}} \right]^{{x}} \:\left[{P}_{{D}} \right]^{{y}} }{\left[{P}_{{A}} \right]^{{m}\:} \left[{P}_{{B}} \right]^{{n}} } \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\left\{\left[{C}\right]{RT}\right\}^{{x}} \:\left\{\left[{D}\right]{RT}\right\}^{{y}} }{\left\{\left[{A}\right]{RT}\right\}^{{m}} \:\left\{\left[{B}\right]{RT}\right]^{{n}} } \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\left[{C}\right]^{{x}\:} \left[{D}\right]^{{y}} \:\left({RT}^{\:} \right)^{{x}+{y}} }{\left[{A}\right]^{{m}} \:\left[{B}\right]^{{n}} \:\left({RT}^{\:} \right)^{{m}+{n}} } \\ $$$$\:\:\:\:\:\:\:\:=\:{K}_{{C}} \:\left({RT}\right) \\ $$$$ \\ $$