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Derive relation between Kc and Kn. (plz answer fast) Derive relation between Kc and Kn. (plz answer fast)
Relationship between Kp and KcConsider the following reversible reaction:aA + bB ⇌ cC + dDThe equilibrium constant for the reaction expressed in terms of the concentration (mol / litre) may be expressed as:K c = [C] c [D] d / [A] a [B] bIf the equilibrium involves gaseous species, then the concentrations may be expressed in terms of partial pressures of the gaseous substance. The equilibrium constant in terms of partial pressures may be given as:K p = pcC pdD / paA pbBWhere pA, pB, pC and pD represents the partial pressures of the substance A, B, C and D respectively. If gases are assumed to be ideal, then according to ideal gas equation:pV = nRTp = nRT / VWhere p ———-> pressure in Pan ——————–> amount of gas in molV ——————–> Volume in m3T ———————> temperature in Kelvinn/V = concentration, Corp = CRT or [gas] RTIf C is in mol dm-3 and p is in bar, then R = 0.0831 bar dm3 mol-1 K-1Therefore, at constant temperature, pressure of the gas P is proportional to its concentration C, i.e.Let us suppose a general reaction:aA + bB↔ cC + dDThe equilibrium constant will be given as:Kp = (pC) c (pD) d /(pA) a (pB) b ……. (1)Now, p = CRTHence,pA = [A] RTwhere [A] is the molar concentration of ASimilarly,pB = [B] RTpC = [C] RTpD = [D] RTwhere [B], [C] and [D] are the molar concentration of B, C and D respectivelySubstituting these values in expression for Kp i.e. in equation (1)Kp = [([C] RT) c ([D] RT) d]/[([A] RT) a ([B] RT) b]= [C] c [D] d (RT) c+d/[A] a [B] b (RT) a+b= [C] c [D] d (RT) c+d – a+b/[A] a [B] b= Kc (RT) c+d – a+b= Kc (RT) ∆nWhere ∆n = (c + d) – (a + b) i.e. number of moles of gaseous products – number of moles of gaseous reactants in the balanced chemical reaction.Hence relation between Kp and Kc is given as:Kp = Kc (RT) ∆n
Relationship between Kp and KcConsider the following reversible reaction:aA + bB ⇌ cC + dDThe equilibrium constant for the reaction expressed in terms of the concentration (mol / litre) may be expressed as:K c = [C] c [D] d / [A] a [B] b
If the equilibrium involves gaseous species, then the concentrations may be expressed in terms of partial pressures of the gaseous substance. The equilibrium constant in terms of partial pressures may be given as:K p = pcC pdD / paA pbBWhere pA, pB, pC and pD represents the partial pressures of the substance A, B, C and D respectively. If gases are assumed to be ideal, then according to ideal gas equation:pV = nRTp = nRT / VWhere p ———-> pressure in Pan ——————–> amount of gas in molV ——————–> Volume in m3T ———————> temperature in Kelvinn/V = concentration, Corp = CRT or [gas] RT
If C is in mol dm-3 and p is in bar, then R = 0.0831 bar dm3 mol-1 K-1
Therefore, at constant temperature, pressure of the gas P is proportional to its concentration C, i.e.
Let us suppose a general reaction:aA + bB↔ cC + dDThe equilibrium constant will be given as:Kp = (pC) c (pD) d /(pA) a (pB) b ……. (1)Now, p = CRTHence,pA = [A] RTwhere [A] is the molar concentration of ASimilarly,pB = [B] RTpC = [C] RTpD = [D] RTwhere [B], [C] and [D] are the molar concentration of B, C and D respectivelySubstituting these values in expression for Kp i.e. in equation (1)
Kp = [([C] RT) c ([D] RT) d]/[([A] RT) a ([B] RT) b]
= [C] c [D] d (RT) c+d/[A] a [B] b (RT) a+b
= [C] c [D] d (RT) c+d – a+b/[A] a [B] b
= Kc (RT) c+d – a+b= Kc (RT) ∆nWhere ∆n = (c + d) – (a + b) i.e. number of moles of gaseous products – number of moles of gaseous reactants in the balanced chemical reaction.
Hence relation between Kp and Kc is given as:Kp = Kc (RT) ∆n
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