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180 chemistry vary as the amount of X in a given volume This shows that at a particular varies. Let us take thermal dissociation of temperature, there is a constant concentration calcium carbonate which is an interesting and or pressure of CO in equilibrium with CaO(s) 2 important example of heterogeneous chemical and CaCO 3(s). Experimentally it has been equilibrium. found that at 1100 K, the pressure of CO 2 in equilibrium with CaCO (s) and CaO(s), is CaCO (s) CaO (s) + CO (g) (6.16) 5 3 3 2 2.0 ×10 Pa. Therefore, equilibrium constant Onthebasisofthestoichiometricequation, at 1100K for the above reaction is: we can write, K p P CO = 2 × 10 Pa/10 Pa = 2.00 ▯CaO s ▯ ▯CO g ▯ 2 K ▯ ▯ 2 ▯ ▯ Similarly, in the equilibrium between c ▯CaCO s ▯ ▯▯ nickel, carbon monoxide and nickel carbonyl 3 Since [CaCO (s)] and [CaO(s)] are both (used in the purification of nickel), 3 constant, therefore modified equilibrium Ni (s) + 4 CO (g) Ni(CO) 4g), constant for the thermal decomposition of the equilibrium constant is written as calcium carbonate will be ▯Ni ▯O ▯ ▯ K´ = [CO (g)] (6.17) K c 4 c 2 ▯O ▯4 or Kp = pCO2 (6.18) It must be remembered that for the existence of heterogeneous equilibrium pure Units of Equilibrium Constant solids or liquids must also be present (however The value of equilibrium constant K can be small the amount may be) at equilibrium, but calculated by substituting the concentration their concentrations or partial pressures do terms in mol/L and for K partial pressure not appear in the expression of the equilibrium p is substituted in Pa, kPa, bar or atm. This constant. In the reaction, results in units of equilibrium constant based Ag 2O(s) + 2HNO (3q) 2AgNO (3q) +H O(2) on molarity or pressure, unless the exponents of both the numerator and denominator are 2 same. ▯AgNO 3 ▯ K c 2 For the reactions, ▯HNO 3▯ H (g) + I (g) 2HI, K and K have no unit.2 2 c p N 2 4g) 2NO 2g), K cas unit mol/L and K has unit bar Problem 6.6 p Equilibrium constants can also be The value of K pfor the reaction, expressed as dimensionless quantities if the CO (g) + C (s) 2CO (g) standard state of reactants and products is 3.0 at 1000 K. If initially P = 0.48 CO2 are specified. For a pure gas, the standard bar and P CO = 0 bar and pure graphite is state is 1bar. Therefore a pressure of 4 bar present, calculate the equilibrium partial in standard state can be expressed as 4 pressures of CO and CO . bar/1 bar = 4, which is a dimensionless 2 number. Standard state (c ) for a solute is 1 Solution 0 molar solution and all concentrations can be For the reaction, measured with respect to it. The numerical value of equilibrium constant depends on the let ‘x’ be the decrease in pressure of CO 2 standard state chosen. Thus, in this system then CO 2g) + C(s) 2CO(g) both Kpand K cre dimensionless quantities Initial but have different numerical values due to different standard states. pressure: 0.48 bar 0 Reprint 2026-27