<p> Chapter 18 Worksheet 3 G under non-standard conditions, relationship between Go and K</p><p>Consider the generalized reaction: aA + bB = cC + dD</p><p>Recall that if you mix A, B, C, and D together, you can decide which direction the reaction will proceed by comparing the reaction quotient (Q) to the equilibrium constant (K) OR by determining the sign of G.</p><p> c d c d [C] [D] [C]eq [D]eq Q  a b K  a b [A] [B] [A]eq [B]eq</p><p>If Q < K or G < 0 then the forward reaction is spontaneous (forward rate > reverse rate). If Q > K or G > 0 then the reverse reaction is spontaneous (forward rate < reverse rate). If Q = K or G = 0 the reaction is at equilibrium (forward rate = reverse rate).</p><p>If the reaction is not at equilibrium, it continues until Q = K and G = 0. Reactions proceed toward minimum free energy just as a ball rolling downhill proceeds toward minimum potential energy.</p><p>Figure legend: The y-axis is G; the x-axis represents the progress of the reaction.</p><p>The sign of G is given by the slope of the curve. When the slope is negative, the forward reaction is “downhill”. When the slope is positive, the reverse reaction is “downhill”. At the minimum, G = 0 and the reaction is at equilibrium.</p><p>This discussion suggests that there should be a mathematical relationship between G and Q.</p><p>1 The relationship between  G and Q: calculating  G under non-standard conditions </p><p>G = Go + RTlnQ (When using this equation, the unit on G must be J/mol or kJ/mol)</p><p>We will not derive this equation, but we can analyze it to see that it makes sense.</p><p>1. If a reaction is set up under standard conditions what is the value of Q and G (use the equation above)?</p><p>Q = ______</p><p>G = ______</p><p>2. As Q increases, what happens to the value of G? Is this as expected?</p><p> o The relationship between  G and Keq and between  G and Q</p><p>3. If a reaction is at equilibrium,</p><p>Q = ______</p><p>G = ______</p><p>4. Insert the values from question 3 into the equation for G above. Rearrange the equation to produce o an expression for G . Rearrange this expression to solve for Keq.</p><p>Go = ______</p><p>eq = ______</p><p>5. Substitute your expression for Go into the original equation for G.</p><p>G = ______</p><p>If Q = Keq, what is the value of G? ______</p><p>If Q < Keq, what is the sign of G? ______</p><p>If Q > Keq, what is the sign of G? ______</p><p>2 o o Relationship between  H ,  S , T, and Keq (expt. 12H)</p><p>Recall that: G = H – TS</p><p>Therefore: Go = Ho – TSo</p><p>From the previous page, there is a second way to calculate Go:</p><p> o G = -RTlnKeq</p><p> o Equate these two expressions for G and solve for lnKeq.</p><p>This is the equation used in experiment 12H! </p><p>A plot of lnKeq vs. 1/T produces a straight line with: slope = –Ho/R y-intercept = So/R (This is called “C” in expt. 12H)</p><p>6. According to the equation above, how does Keq depend on temperature for an endothermic or exothermic reaction?</p><p>3 7. Consider the reaction 2NO2 (g) → N2O4 (g)</p><p> a. Is this reaction spontaneous under standard conditions at 298 K? At 350 K? Answer these </p><p> questions in two ways: determine the sign of G and compare Q to Keq.</p><p>4 b. Is the reaction spontaneous at 350 K if the reaction mixture consists of 1.60 atm of NO2 and 0.40 atm of N2O4? Answer in two ways as in part a.</p><p>5 Summary</p><p>G and Q provide the same information: How far the reaction is from equilibrium and which direction the reaction must proceed in order to reach equilibrium. The further a reaction is from equilibrium, the more work it can do.</p><p> o G and Keq provide the same information: The concentrations of reactants and products present at equilibrium.</p><p>The second law of thermodynamics explains how Hrxn and Srxn determine the direction in which a reaction will proceed.</p><p>Thermodynamics explains the relationship between temperature and Keq</p><p>Thermodynamics provides a way to use calorimetry to measure equilibrium constants.</p><p>6</p>