Enduring Understanding 6.A: Equilibrium Problems

  • Sample Question 1: A chemical reaction will always be spontaneous (thermodynamically favored) under which conditions?
  • a) Exothermic, with an increase in entropy
    b) Exothermic, with a decrease in entropy
    c) Endothermic, with a decrease in entropy
    d) Endothermic, with an increase in entropy
  • Answer: (a), Exothermic with an increase in entropy
  • Thermodynamic favorability of a chemical reaction is determined by the Gibbs free energy,
  • ΔG = ΔH - TΔS
  • If ΔG < 0, the reaction is spontaneous.
  • Therefore, if ΔH is negative (reaction is exothermic) and ΔS is positive (increase in entropy) then ΔG must always be negative.
  • Sample Question 2: Consider the reaction X(g) + Y(g) ⇆ Z(g), with K = 10 at 25 °C
  • Sample [X][Y] [Z]
    10.5210
    240.2510
    34120
    4115
  • Which of the above samples are at equilibrium?
  • a) Sample 1
    b) Sample 2
    c) Samples 1 and 2
    d) Samples 1, 2, and 4
  • Answer: (c), Samples 1 and 2
  • The equilibrium constant is determined by K=[Z]/[X][Y]. If [Z]/[X][Y] = 10, the sample is at equilibrium. In both samples 1 and 2, [Z]/[X][Y] is 10, so they are both at equilibrium. In samples 3 and 4, [Z]/[X][Y] is 5, and the samples are not at equilibrium.
  • Sample Question 3: Consider the reaction 2 NO2(g) ⇆ N2O4 (g). A rigid container is pressurized with 10 atm NO2, then allowed to come to equilibrium with a final concentration of 4.0 atm N2O4. What is Kp?
  • a) 1
    b) 2
    c) 4
    d) 10
  • Answer: (a), 1
  • The problem can be solved as follows:
  • Two equivalents NO2 are used to form 1 equivalent N2O4, so at 4.0 atm N2O4 the amount of NO2 remaining will be (10 - 2 x 4.0) or 2.0 atm.
  • CompoundNO2N2O4
    Initial pressure (atm)100
    Final Pressure (atm)2.04.0
  • So the equation for K is
  • Kp = [N2O4] / [NO2]2 4.0 / (2.0)2 = 4.0 / 4.0 = 1.0



Related Links:
Chemistry
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AP Chemistry Notes
Reversible Reactions



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