(b) The Equilibrium Constant and the Reaction QuotientThe reaction quotient and the equilibrium constant for a reaction have the same expression, but the numerical values can be different. The reaction quotient is, in fact, more general, because Qtakes values for any conditions. Ktakes values specified by the thermodynamics of the process. Now, we can relate ∆G°to the equilibrium constant Kand ∆Gto the reaction quotient Qby recognizing that when ∆G< 0; Q< Kand when ∆G= 0, Q= K. Therefore, we have the following equation: ∆G(T) = ∆G°(T) + RTln Q= −RTln K + RTln Q= lnQRTK. General Standard Conditions Conditions where R= 8.314 J/K mol and Tis in degrees Kelvin. To summarize equilibrium then: Sign of ∆GSize of Qvs. KSpontaneous? −Q < K Yes (left to right) 0 Q = K Equilibrium (quasistatic) + Q > K No (left to right) Example: How can we use Qand K? Consider the following homogeneous equilibrium: 2 NO(g) + O2(g) ZZXYZZ2 NO2(g) at T =750 K ∆H0= −113.1 kJ ∆S0= −145.3 J/K ∆G0(750 K) = −4.13 kJ 222NO2NOO(750 K)1.94ppKpp==2 NO2(g) ZZXYZZN2O4(g) is another, simultaneous equilibriumthat also occurs. 242N O62NO' (750 K)6.5510ppKp−==×Consider the following conditions: NOp2Op2NOpQ vs. Kp ∆GComments / Impact on 24N Op1.00 atm 1.00 atm 0 atm Q= 0 < Kp< 0NO2forms; also forms N2O42.00 atm 1.00 atm 2.00 atm Q= 1 < Kp< 0NO2forms; also forms N2O41.00 atm 1.00 atm 4.00 atm Q= 16 > Kp> 0NO2consumed; also reduces N2O4
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