Chemistry Week 13

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
(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) Z ZX 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 pp == 2 NO 2 (g) ZZX N 2 O 4 (g) is another, simultaneous equilibrium that also occurs. 24 2 NO 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 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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2.00 atm 0 atm 0 atm Undefined: not an equilibrium situation 0 atm 0 atm 1.00 atm Q = > K p > 0 NO 2 consumed; also reduces N 2 O 4 1.605 atm 0.2 atm 1.00 atm Q = 1.94 = K p = 0 24 NO p = 6.55 × 10 6 atm Now, increase the pressure by reducing the volume by 50% (1/2): 3.21 atm 0.4 atm 2 atm Q = 0.97 < K p < 0 NO 2 forms (c) LeChâtelier’s Principle : “If a system at equilibrium is disturbed by a change in temperature, pressure or concentration of one of the components, the system will respond to re-establish equilibrium by counteracting the effect of the disturbance.” Example : 2 NO(g) + O 2 (g) ZZX YZZ 2 NO 2 (g) + Heat H 0 = 113.1 kJ (1) Add O 2 (g) – increases 2 O p , so that Q < K p . NO 2 forms. (2) Increase total pressure by reducing the volume – Q < K p . NO 2 forms – to re-establish equilibrium, the reaction proceeds that reduces the total amount of gas. (3) Increase temperature – NO 2 consumed. NOTE: K p decreases with increasing T . Example : 2 NO 2 (g) N 2 O 4 (g) + Heat H 0 = 58.05 kJ NO 2 (g) is a brown gas; N 2 O 4 (g) is a colorless gas. This equilibrium is exothermic: if you examine the Lewis structures, NO 2 has one unpaired electron at the N atom, N 2 O 4 has a N N single bond. Therefore, H 0 D (N N) = 163 kJ/mol. The actual enthalpy difference is much less, which can be attributed to intermediate distance O···O repulsions. At low temperatures, the right-hand side is preferred (colorless); at high temperatures, the left-hand side is preferred (brown).
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/27/2008 for the course CHEM 201 taught by Professor Miller during the Fall '07 term at Iowa State.

Page1 / 12

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online