Chem 116 POGIL Worksheet - Week 7 Kinetics to Equilibrium - Solutions 2 NO 4 NO + O. Observed Rate = k[no] 2 5

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1 Chem 116 POGIL Worksheet - Week 7 Kinetics to Equilibrium - Solutions Key Questions 1. Consider the gas phase reaction 2 NO NO 2 + O2 for which the observed rate law expression is Observed Rate = k[no] 2 5 The following mechanism has been proposed: NO 2 5 NO 2 + NO k 1 ( ), k 1 ( ) fast equilibrium NO 2 + NO NO + NO 2 + O2 k2 slow NO + NO 2NO k fast 2 Carry out the following steps to show that this is a plausible mechanism. i. Show that this series of steps adds to give the overall stoichiometry of the reaction. [Hint: One step needs to be multiplied by an integer factor.] 2(N2O 5 NO 2 + NO ) NO 2 + NO NO + NO 2 + O2 NO + NO 2 NO2 2 NO 4 NO + O ii. Write the rate law for each step. The first step is reversible, so you will need to write two rate law expressions, one for each direction. Step 1, forward: rate 1 = k 1[NO] 2 5 Step 1, reverse: rate 1 = k 1[NO 2][NO ] Step 2: rate 2 = k 2[NO 2][NO ] Step : rate = k [NO ][NO] iii. Which step is rate determining. Write its rate law expression. Step 2 is rate determining, for which rate 2 = k 2[NO 2][NO ] iv. What species are reaction intermediates? NO and NO

2 v. Your rate expression, based on the rate expression for the rate-determining step, contains a concentration term for an unobservable reaction intermediate. We need to express the rate law for the rate determining step in terms of observable starting materials. Note that the first step is a rapidly established equilibrium, which means its rate in the forward direction is exactly the same is in the reverse direction. Write an equality between your two expressions for the two directions, and rearrange it to solve for [NO ]. k[no] = k -1[NO 2][NO ] Solving for [NO ] gives vi. Substitute your expression for [NO ] into your expression for the rate law of the rate determining step. Does this agree with the experimentally observed rate law expression? This is the same as the experimentally observed rate expression, if k exp = k1k/k 2 1. On this basis, the proposed mechanism is plausible. 2. Consider the following reversible reaction, which is believed to proceed by a one-step mechanism in each direction: a. Write the rate expression for each direction. rate = k [NO ] rate = k [NO] 2 f f 2 r r 2 4 b. At equilibrium, the net rate of the reaction is zero, because the rate of the forward reaction is exactly equal to that of the reverse reaction. Set your two rate expressions equal to each other, and solve the equality for k/k. f r This ratio of constants is itself a constant, which we call the equilibrium constant, K. c 2 k[no f 2] = k r[no] 2 4

3 . Write the expressions for the equilibrium constant, K c for the following reactions. a. N 2(g) + H 2(g) 2NH (g) b. Cu(NO ) 2(aq) + Zn(s) Cu(s) + Zn(NO ) 2(aq) [Hint: Write the net ionic equation first.] Cu (aq) + Zn(s) Cu(s) + Zn (aq) Note that the spectator ions do not appear in the K c expression c. CaO(s) + CO 2(g) CaCO (s) Note that the solids are omitted from the K c expression. d. CaCO (s) CaO(s) + CO 2(g) How does this compare to the expression you wrote for the reverse reaction in part c? K c = [CO 2] This is the inverse of the expression for the reverse reaction. e. HF(aq) + H2O(l) HO(aq) + F (aq) [Hint: H2O(l) is a solvent.]

4 4. Consider the equilibrium HF(aq) + H2O(l) HO(aq) + F (aq) and its K c as you defined it in Key Question.e. A solution of hydrofluoric acid is prepared by dissolving mol HF in enough water to make a liter of solution. Once equilibrium is established, the concentration of hydronium ion is found to be [H O ] = 7.91 x 10 M. a. What is the concentration of undissociated HF in the solution? To answer this question, fill in the values in the following table. [HF] [HO ] [F ] Start M ~0 M 0 M Change 7.91 x 10 M x 10 M x 10 M Equilibrium M x 10 M x 10 M b. What is the value of K c for HF? c. What is the concentration of undissociated HF in a hydrofluoric acid solution in which [H O ] = [F ] = M? d. How many moles per liter of HF were added to make the solution in part c? Let C HF be the number of moles per liter of HF added to make the solution. At equilibrium, every mole of F that exists in the solution must have required the loss of the same number of moles of HF. Therefore, at equilibrium, we could write [HF] = C HF [F ] From part c we have numerical values for [HF] and [F ], so we can rearrange this expression to solve for C : HF C HF = [HF] + [F ] = M M = M

5 5. At 1000 K, the value of K c is 4.08 x 10 mol/l for the equilibrium 2 SO (g) 2 SO 2(g) + O 2(g) What is the value of K p? [R = L atm/k mol] For this reaction, n = 2 = 1. Therefore, K = K RT. K p= (4.08 x 10 mol/l)( L atm/k mol)(1000k) = 0.5 atm 1 6. At 800K, the value of K p is 4.55 atm for the equilibrium What is the value of K c? CaO(s) + CO 2(g) CaCO (s) For this reaction, counting only moles of gas species, n = 0 1 = K p = K(RT) c K c = KpRT = (4.55 atm )( L atm/k mol)(800k) = 00 L/mol p c