Dr. Zellmer Chemistry 1220 Monday Time: 18 mins Spring Semester 2019 February 4, 2019 Quiz III. Name KEY Rec. TA/time

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1 Dr. Zellmer Chemistry 1220 Monday Time: 18 mins Spring Semester 2019 February 4, 2019 Quiz III Name KEY Rec. TA/time 1. (8 pts) For the following reaction K P = at EC. 2 Cl 2 (g) + 2 H 2 O (g) W 4 HCl (g) + O 2 (g) Rxn 1 (1) a) (3 pts) What is the value of K P for the following reaction? Show all work or explain. 6 HCl (g) + 3/2 O 2 (g) W 3 Cl 2 (g) + 3 H 2 O (g) (2) When multiply a rxn by some factor the K for the new rxn is equal to original K raised to the factor the rxn was multiplied by. Remember if you multiply a reaction by a -1 that s the same as reversing it since you need to flip the action around to get rid of the negative coefficients you would get by multiplying by -1. (multiplying by -1 reverses the rxn) Based on this K rev for rev. rx is 1/K for (K for!1, the reciprocal of K for for the forward rx) To get reaction (2) you need to reverse the original reaction (rxn 1) and multiply it by 3/2. This is the same as multiplying reaction (1) by - 3/2. K 2 = K 1!3/2 = 1/(K 1 3/2 ) = (0.0752)!3/2 = = 48.5 (3 s.f.) b) (3 pts) What is the value of K C for the original Rxn 1 at EC? Show all work or explain. K p = K c (RT) n Δn = change in moles of gas Do NOT include pure solids or liquids when calculating Δn. = (moles of gaseous products)! (moles of gaseous reactants) R = LCatm/molCK NOT J/molCK T = EC = K (must be in kelvin) K P = Δn = 5-4 = 1 K C = K p (RT)! n K C = K P (RT)!(1) = (0.0752){( ) (753.15)}!1 K C = x 10!3 = 1.22 x 10!3 (3 s.f.) ***** continued on next page *****

2 1. (Cont.) c) (2 pts) What does the size of K P for Rxn 1 mean about the amounts of reactants and products at equilibrium? (P HCl ) 4 (P O2 ) K P = = (P Cl2 ) 2 (P H2O ) 2 appreciable amounts of reactants & products at equil. Very large K c (K c > ) : Very small K c (K c > 10!10 ) : Large K c (10 2 < K c < ) : Small K c (10!10 < K c > 10!2 ) : essentially only products (at equil.) - for all practical purposes essentially only reactants (at equil.) - for all practical purposes products predominate reactants predominate 0.01 < K c < 100 : significant amounts of both reactants & products present at equil. (Roughly equal molar amounts depending on the form of K c.) Look at the following example: A + B W X + Y [X] [Y] K c = [A] [B] Both numerator and denominator are squared (have same form). So if K. 1 then there are roughly equal molar amounts of reactants and products. Look at ICE (equilibrium) tables starting with 10 3 M conc. of A & B (rather large conc.): If K c = 1 x 10!10 you will get the following conc. for reactants & products at equil. [A] = [B] = M and [X] = [Y] = 1.0 x 10!5 M. Certainly reactants predominate but these. conc for [X] and [Y] are still very easily measurable. If K c = 1 x 10!20 you will get the following conc. for reactants & products at equil. [A] = [B] = M and [X] = [Y] = 1.0 x 10!10 M. These conc. for [X] and [Y] are much harder to measure and there are essentially only reactants at equilibrium. Look at equilibrium tables starting with 1.0 M conc. of A & B (much more reasonable starting conc.) and the K c given in the problem: If K c = 9.5 x 10!33 you will get the following conc. for reactants & products at equil. [A] = [B] = M and [X] = [Y] = 9.7 x 10!17 M. These conc. for [X] and [Y] are very difficult to measure and there are essentially only reactants. Even for a K c = 1 x 10!10 you will get the following conc. for reactants & products at equil. [A] = [B] = M and [X] = [Y] = 1 x 10!5 M. Certainly reactants predominate but these conc. for [X] and [Y] are still easily measurable. I said the following in class. A really large K (around ) - essentially only products (as is the case for strong acids). A really small K (around 10!15 ) - essentially only reactants. A large K (> 10 2 ) - mostly products (products predominate). A small K (> 10!2 ) - mostly reactants (reactants predominate). If 0.01 < K c < significant (appreciable) amounts of both reactants & products present at equil.

3 2. (6 pts) The following reaction is started with moles of BrCl in a L container. When equilibrium is reached there are moles of BrCl in the container. What is the equilibrium constant, K c? (Show the ICE table. When appropriate, state any assumptions made and check your percent error.) 2 BrCl (g) W Br 2 (g) + Cl 2 (g) [Br 2 ] C [Cl 2 ] K C = [BrCl] 2 Concentration of Products in Numerator Concentration of Reactants in Denominator Use coefficients in balanced equation as the exponents of the concentrations. Pure solids (s) and liquids (R) do not appear in the expression for K or Q. The solution to the actual problem given is: Use ICE (equil.) table (in in Molarity when dealing w. K C ) you know rxn must proceed 2 BrCl (g) W Br 2 (g) + Cl 2 (g) to right (in forward direction) initial M 0 M 0 M since starting with no Br 2 or change -2x + x + x Cl 2. Conc. of BrCl is M since you have mol/2 L. equil x x x Must use Molarity in ICE table. [Br 2 ] [Cl 2 ] (x) (x) x 2 K = = = [BrCl] 2 ( x) 2 ( x) 2 Can solve for x from the information given for the equilibrium concentration of BrCl. ( x) mol/2.000 L = M x = M (2 s.f. due to subtraction) x 2 (0.0215) 2 K C = = = = 0.14 ( x) 2 { (0.0215)} 2 (2 s.f.)

4 3. (2 pts) Using concentrations, write the equilibrium constant expression, K C, for the following reaction. Fe (s) + 2 H + (aq) W Fe 2+ (aq) + H 2 (g) [H 2 ] [Fe 2+ ] K c = [H + ] 2 Concentration of Products in Numerator Concentration of Reactants in Denominator Use coefficients in balanced equation as the exponents of the concentrations. Pure solids (s) and liquids (R) do not appear in the expression for K or Q. Fe (s) is pure solid - pure solids and liquids do NOT appear in the equilibrium constant expression, whether it s K P or K C. 4. (6 pts) For the following reaction K c equals x 10 3 at 218 EC, what are the equilibrium concentrations of NH 3 and H 2 S if mole of each is placed in a L vessel and allowed to react until equilibrium is reached? (Show the ICE table. When appropriate, state any assumptions made and check your percent error.) Use ICE (equil.) table (in Molarity when dealing w. K C ) NH 3 (g) + H 2 S (g) W NH 4 HS (s) initial 2.00 M 2.00 M 0 change - x - x + x equil x x + x 1 K C = = x 10 3 [NH 3 ] C [H 2 S] 1/( x) 2 = x 10 3 NH 4 HS (s) doesn t appear in K (pure solids and liquids do not appear in K) x = (1.20 x 10-4 ) 1/2 = x 10-2 M (x = 1.989) [NH 3 ] = [H 2 S] = 1.10 x 10-2 M

5 5. (3 pts) A plot of ln(r) vs. ln[a] has a slope of 5 and an intercept of Determine the rate constant and order of the reaction for the rate law, r = k[a] n. r = k [A] n k = rate constant n = order Convert this to a linear equation. Take the log of both sides, use the rules of logarithms and rearrange to make it look like an equation for a straight line. You can use either log or ln. log(r) = log{k[a] n } = log(k) + log{[a] n } = log(k) + n log[a] log(r) = n*log[a] + log(k) or ln(r) = n*ln[a] + ln(k) y = m*x + b Using the equation from (a) how do you graphically obtain k and n (i.e. what do you plot as x and y and what are n and k related to on the graph)? The data you have is concentration, [A], and the corresponding rate. You take the log of each and plot log(r) vs log[a] (i.e. log(r) is on the y axis and log[a] is on the x axis) or ln(r) and ln[a]. if using log: log(k) = y-intercept k = 10 (y-int) if using ln: ln(k) = y-intercept k = e (y-int) n = slope (The slope can be either positive or negative depending on the order.) n = 5 ln(k) = &7.809 k = e &7.809 = 4.06 x 10 &4 M -4 /s