K c = [NH 3 ] C [H 2 S] = x 10!4

Transcription

1 Dr. Zellmer Chemistry 1250 Wednesday Time: 30 mins Spring Semester 2019 April 17, 2019 Quiz III Name KEY Lab TA/time 1. (10 pts) For the following reaction K c equals x 10!4 at 218EC. NH 4 HS (s) W NH 3 (g) + H 2 S (g) a) (5 pts) What are the equilibrium concentrations of NH 3 and H 2 S if solid NH 4 HS is placed in a closed container and decomposes until equilibrium is reached at 218EC? (Show the ICE table. State any assumptions made and check your percent error.) Use ICE (equil.) table (in Molarity) NH 4 HS (s) W NH 3 (g) + H 2 S (g) initial change - x - x + x equil x + x K c = [NH 3 ] C [H 2 S] = x 10!4 x 2 = x 10!4 NH 4 HS (s) doesn t appear in K (pure solids and liquids do not appear in K) x = (1.200 x 10!4 ) 1/2 = x 10!2 M [NH 3 ] = [H 2 S] = x 10!2 M b) (1 pt) What happens to the reaction when the pressure is increased (by decreasing the volume of the container) at constant temperature? (i.e. does the equilibrium shift and if so in what direction? If no shift then why not.) EXPLAIN! 0 mol gas 2 mol gas NH 4 HS (s) W NH 3 (g) + H 2 S (g) Increase pressure (by decreasing volume): Shifts to left (0 mol gas <--- 2 mol gas) Increase pressure (by decreasing volume) : shift to the side with fewer moles of gas. For Pressure changes (due to volume changes): P inc ==> fewer moles gas P dec ==> more moles gas Changes in Pressure do not affect liquids, solids or aqueous (their conc. do not change w. P like conc. of gases do) If no change in moles gas changes in P have no effect

2 1. (Continued) c) (1 pt) What happens to the reaction when NH 3 is removed? (i.e. does the equilibrium shift and if so in what direction? If no shift then why not.) EXPLAIN! Shifts to the right (forward direction) to replace some of the removed substance. When you remove something (other than a pure solid or liquid) the reaction shifts toward the removed substance to replace it. Reactant Add Product Remove Shift Right Remove Add Left Move AWAY from what is ADDED Move TOWARD what is REMOVED NOTE: K remains constant for changes in conc. & pressure (Numerical value of K changes only for temp. changes NOTE: Adding or Removing Pure SOLIDS of LIQUIDS (as long as some is present) does NOT affect the equil. This is because the conc. of pure solids and liquids are essentially constant and do NOT appear in K. d) (3 pts) What is the value of K p for this reaction? K p = K c (RT) n (Δn = change in moles of gas, products - reactants) Δn = 2! 0 = 2 Do NOT include the solid (if you do Δn = 1) R = LCatm/molCK T = 218EC = 491 K NOT J/molCK (must be in kelvin) K p = x 10!4 {( )(491 K)} 2 = = 0.195

3 2. (2 pts) Which of the following are strong acids or strong bases? (Circle all that apply.) HNO 3 H 3 PO 4 HClO 3 CsOH NH 3 HBrO 3 H & HSO 4 & HF C 6 H 5 OH The 7 strong acids are: HCl, HBr, HI binary acids HNO 3, HClO 3, HClO 4, H 2 SO 4 (1 st H + only) oxyacids (ternary) Strong bases are all group 1 A and soluble group 2 A (Ca and below) hydroxides & bases stronger than OH & (some examples are below) O 2& (conj base of OH & ), H & (conj base of H 2 - a neutral molecule), NH 2 & (conj base of NH 3, itself a base), NH 2& (conj base of NH 2 & ), N 3& (conj base of NH 2& ), S 2& (conj base of SH & itself the conj base of H 2 S), CH 3 & (conj base of CH 4, a neutral nonionizing molecule in H 2 O), CH 3 O & (conj base of CH 3 OH, a neutral nonionizing molecule in H 2 O) 3. (2 pts) What is(are) the difference(s) between the Bronsted-Lowry and Lewis definitions of an acid? Arrhenius acid: A substance that has a hydrogen, H, in the formula and when dissolved in water, increases the concentration of H + ions. Restricted to aqueous solutions. Bronsted-Lowry acid: A substance (molecule or ion) that can donate a proton, H +, to another substance. This theory is about proton transfer reactions. It is NOT restricted to aqueous solutions, whereas Arrhenius theory is. This theory also introduced the idea about conjugate acid-base pairs. Has to have a H to donate. Proton Donor This theory covers all Arrhenius acids and bases. Lewis acid: A substance (molecule or ion) that can accept a pair of electrons from another substance. It is NOT restricted to aqueous solutions, whereas Arrhenius theory is. Electron-pair Acceptor This is the most comprehensive theory of the three. It covers all B-L acids and bases (and Arrhenius acids and bases) and reactions which can explain why things w/o hydrogens in the formula can cause a solution to be acidic.

4 4. (3 pts) Would you expect the following solutions to be acidic, neutral, or basic? Explain! In general: cations => acidic soln (except group 1A & 2A(Ca, Sr, Br) cations - give neutral) anions => basic soln (except conj. bases of strong acids; Cl -, Br -, I -, NO 3 -, ClO 3 -, ClO 4 - are neutral HSO 4 - is acidic) H 2 O a) KClO > K + (aq) + ClO 3! (aq) NEUTRAL K + : group 1A cation ( conj. acid of a strong base, KOH, and does not act as an acid) - neutral ClO 3! is conj. base of a strong acid, HClO 3, & does not act as a base - neutral H 2 O b) AlCl > Al 3+ (aq) + 3 Cl! (aq) ACIDIC Cl! : conj. base of a strong acid, HCl & does not act as a base - neutral Al 3+ : metal cation - causes soln to be acidic

5 5. (12 pts) A solution of M benzoic acid, HC 7 H 5 O 2, is 3.5% ionized. (Show the ICE table. State any assumptions made and check your percent error.) Show work and explain! a) What are [H + ], [OH! ], ph and poh in this solution? [H + ] % ionization = x 100% [HC 3 H 5 O 2 ] [H + ] 3.5 % = x 100% [H + ] = 1.75 x 10!3 M = 1.8 x 10!3 M ph =!log [H + ] =!log (1.75 x 10!3 ) = = 2.76 (actually only 2 s.f. - to right of decimal) ph + poh = pkw = [H + ] x [OH! ] = Kw = 1.0 x 10!14 poh = = (2 s.f.) 1.0 x 10!14 or [OH - ] = = x 10!12 [OH & ] = 10!pOH = 10!11.24 = 5.7 x 10! x 10!3 poh = - log (5.714 x 10!12 ) = b) What is the K a for HC 7 H 5 O 2? (Show the ICE table.) HC 7 H 5 O 2 is a weak acid - weak acid equil. problem - Use ICE (equil.) table (in Molarity) HC 7 H 5 O 2 W H + + C 7 H 5 O 2! (or HC 7 H 5 O 2 + H 2 O W C 7 H 5 O 2! + H 3 O + ) initial change - x + x + x equil & x x x x = [H + ] = 1.75 x 10!3 [C 7 H 5 O 2 & ] [H + ] x 2 (1.75 x 10!3 ) 2 K a = = = = x 10!5 = 6.3 x 10!5 [HC 7 H 5 O 2 ] & x (0.050 & 1.75 x 10!3 )

6 6. (3 pts) Predict the sign of ΔS of the system or if it s approximately zero for the following reactions and explain your choices. a) 2 NH 3 (g) v N 2 (g) + 3 H 2 (g) ΔS > 0 (+) The number of moles of gas is increasing. Generally speaking, when the number of moles of gas increases in a reaction that leads to an increase in entropy and ΔS is positive. (If the change in moles of gas is small one also has to consider the molecules themselves and the number of degrees of freedom in those molecules). ΔS = J/K (calc. from SE values) b) Mg(OH) 2 (s) + 2 HCl (g) v MgCl 2 (s) + 2 H 2 O (g) ΔS. 0 (+) The number of moles of gas is on each side is the same and there is solid on each side. In this case I would expect someone to say the change in entropy is small and approx. zero. It would likely be a positive number (inc. in entropy) due to the fact the bent H 2 O molecule has many more rotational and vibrational degrees of freedom than the linear HCl and thus H 2 O many more microstates. ΔS = J/K (calc. from SE values)

7 7. (8 pts) Given ΔHE = kj and ΔSE = J/molCK for the following reaction, N 2 F 4 (g) v 2 NF 2 (g) a) (3 pt) Calculate the ΔGE of the reaction at 25EC. Is the reaction spontaneous or nonspontaneous at this temperature under standard state conditions? Show all work and explain. ΔGE = ΔHE & T ΔSE = 85.0 kj & ( K)( kj/k) = kj T in kelvin; ΔSE converted to kj/k = kj ΔGE > 0 NONspontaneous under standard conditions (Nonspont. to products - reactans predominate at equil.) b) (3 pts) If the reaction is nonspontaneous, at what temperature would it be spontaneous. If the reaction is spontaneous, at what temperature would it be nonspontaneous. If the reaction will always be spontaneous at all temperatures or never be spontaneous at any temperature state that. Rxn. is NONspont. at T = 25EC. At what temp will it become spont., i.e. at what temp will ΔGE < 0 ΔHE > 0 and ΔSE > 0 so rxn is spont at high temp ΔGE = ΔHE & T ΔSE < 0 ΔHE < T ΔSE 85.0 kj < T ( kj/k) 85.0 kj < T (Remember, when multiply or divide by a negative number an inequality ( kj/k) (<, >) flips (>, <) K < T T > 156EC (becomes spont. at temp above this) (becomes more spont as temp inc) c) (2 pt) What is the equilibrium constant at 25EC? ΔGE = & RT lnk ln K = & ΔGE/RT = & ( x 10 4 J)/(8.314 J/K)( K) = & (3 s.f.) K = e! = 2.82 x 10!5 (fairly small but not extremely small - mostly reactants at equilibrium, but not only reactants)