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Equilibrium

Question
CBSEENCH11006365

Derive the relation between Kp and Kc.

Solution
Consider a gaseous reaction in a state of equilibrium.
aA plus bB space rightwards harpoon over leftwards harpoon space space cC left parenthesis straight g right parenthesis space plus space dD left parenthesis straight g right parenthesis
Let PA, PB, Pc and pD be the partial pressures of A, B, C and D respectively at equilibrium, then
          straight K subscript straight p space equals space fraction numerator straight p subscript straight C space end subscript superscript straight c straight p subscript straight D superscript straight d over denominator straight p subscript straight A superscript straight a space cross times space straight p subscript straight B superscript straight b end fraction space space space space space space... left parenthesis 1 right parenthesis
and  
  straight K subscript straight c space equals space fraction numerator open square brackets straight C close square brackets to the power of straight c space open square brackets straight D close square brackets to the power of straight d over denominator open square brackets straight A close square brackets to the power of straight a space open square brackets straight B close square brackets to the power of straight b end fraction    ...(2)
We know that for an ideal gas,
                   PV = nRT
or space space straight P space equals space straight n over straight V RT space equals space Molar space concentration space cross times space space RT
therefore space space space space space space space space straight p subscript straight A space equals space open square brackets straight A close square brackets space space RT
space space space space space space space space space space space straight p subscript straight B space equals open square brackets straight B close square brackets space RT
space space space space space space space space space space straight p subscript straight C space equals space open square brackets straight C close square brackets space RT
space space space space space space space space space space straight p subscript straight D space equals open square brackets straight D close square brackets space RT
Substituting the value of partial pressure in equation (1),
  straight K subscript straight p space equals space fraction numerator open square brackets straight C close square brackets to the power of straight c space left parenthesis RT right parenthesis to the power of straight c space open square brackets straight D close square brackets to the power of straight d space left parenthesis RT right parenthesis to the power of straight d over denominator open square brackets straight A close square brackets to the power of straight a space left parenthesis RT right parenthesis to the power of straight a space open square brackets straight B close square brackets to the power of straight b space left parenthesis RT right parenthesis to the power of straight b end fraction
space space space space equals fraction numerator open square brackets straight C close square brackets to the power of straight c space open square brackets straight D close square brackets to the power of straight d space left parenthesis RT right parenthesis to the power of straight c plus straight d end exponent space over denominator open square brackets straight A close square brackets to the power of straight a space open square brackets straight B close square brackets to the power of straight b space left parenthesis RT right parenthesis to the power of straight a plus straight b end exponent end fraction
therefore space space space straight K subscript straight p space equals space straight K subscript straight c left parenthesis RT right parenthesis to the power of left parenthesis straight c plus straight d right parenthesis minus left parenthesis straight a plus straight b right parenthesis end exponent
                                             
[From equation (2)]
   or space space space straight K subscript straight p space equals space straight K subscript straight c left parenthesis RT right parenthesis to the power of increment straight n end exponent
where space increment straight n space equals space left parenthesis Total space number space of space moles space of space gaseous space products right parenthesis
space space space space space space space space space space minus left parenthesis Total space No. space of space moles space of space gaseous space reactants right parenthesis
straight R space equals space Gas space constant space left parenthesis 0.831 space litre space bar space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent right parenthesis
straight T space equals space Kelvin space temperature