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when gaseous NO and NO2(nitrogen dioxide) are mixed,the following euilibra are established..1.2NO2(g)---N2O4(g) Kp= 6.82. NO(g) = NO2(g)----N2O3(g)in an experiment NO and NO2 are in molar ratio of 1:2 , the final pressure was 5.5 bar and the partial pressure of N2O4(dinitrogen pentaoxide) was 1.7 bar.calculate a) the euilibrium partial pressure of NO and b) the euilibrium constant of reaction 2.....
Dear student,
Consider the following reversible reaction: aA + bB ? cC + dD The 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 pbB Where 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 = nRT p = nRT / V Where p ———-> pressure in Pa n ——————–> amount of gas in mol V ——————–> Volume in m3 T ———————> temperature in Kelvin n/V = concentration, C or p = 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 + dD The equilibrium constant will be given as: Kp = (pC) c (pD) d / (pA) a (pB) b ……. (1) Now, p = CRT Hence, pA = [A] RT where [A] is the molar concentration of A Similarly, pB = [B] RT pC = [C] RT pD = [D] RT where [B], [C] and [D] are the molar concentration of B, C and D respectively Substituting 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) ?n
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Sagar Singh
B.Tech, IIT Delhi
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