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Chemistry Week 13 - (b The Equilibrium Constant and the...

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(b) The Equilibrium Constant and the Reaction Quotient The 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 Q takes values for any conditions. K takes values specified by the thermodynamics of the process. Now, we can relate G ° to the equilibrium constant K and G to the reaction quotient Q by recognizing that when G < 0; Q < K and when G = 0, Q = K . Therefore, we have the following equation: G ( T ) = G ° ( T ) + RT ln Q = RT ln K + RT ln Q = ln Q RT K . General Standard Conditions Conditions where R = 8.314 J/K mol and T is in degrees Kelvin. To summarize equilibrium then: Sign of G Size of Q vs. K Spontaneous? Q < K Yes (left to right) 0 Q = K Equilibrium (quasistatic) + Q > K No (left to right) Example : How can we use Q and K ? Consider the following homogeneous equilibrium : 2 NO(g) + O 2 (g) ZZX YZZ 2 NO 2 (g) at T = 750 K H 0 = 113.1 kJ S 0 = 145.3 J/K G 0 (750 K) = 4.13 kJ 2 2 2 NO 2 NO O (750 K) 1.94 p p K p p = = 2 NO 2 (g) ZZX YZZ N 2 O 4 (g) is another, simultaneous equilibrium that also occurs. 2 4 2 N O 6 2 NO ' (750 K) 6.55 10 p p K p = = × Consider the following conditions: NO p 2 O p 2 NO p Q vs. K p G Comments / Impact on 2 4 N O p 1.00 atm 1.00 atm 0 atm Q = 0 < K p < 0 NO 2 forms; also forms N 2 O 4 2.00 atm 1.00 atm 2.00 atm Q = 1 < K p < 0 NO 2 forms; also forms N 2 O 4 1.00 atm 1.00 atm 4.00 atm Q = 16 > K p > 0 NO 2 consumed; also reduces N 2 O 4
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