HW 13 Q1 What are the units of Kp and Kc for each of the following? (a) NH4S(s) Kp atm2 K c M2 (b) P4(g) + 5 O2(g) Kp atm-6 Kc M-6 NH3(g) + H2S(g) 𝑲𝑲 = [𝑵𝑵𝑯𝑯𝟑𝟑 ][𝑯𝑯𝟐𝟐 𝑺𝑺] 𝑲𝑲 = Q2 P4O10(s) 𝟏𝟏 [𝑷𝑷𝟒𝟒 ][𝑶𝑶𝟐𝟐 ]𝟓𝟓 Consider the following reaction: 2 H2S(g) 2 H2(g) + S2(g) Kp was measured to be 0.0365 atm for this reaction at 385 K. What is the value of Kc at 385 K? 0.00115 Kc = 𝐾𝐾𝑐𝑐 = M �𝐾𝐾𝑝𝑝 � 0.0365 = = 0.00115 𝑀𝑀 (𝑅𝑅𝑅𝑅)∆𝑛𝑛 𝑔𝑔 �0.0821(385)�1 What is the value for Kp for the following reaction at 385 K? H2(g) + 1/2 S2(g) Kp = 5.23 H2S(g) atm-1/2 The reaction is reverse and half of the starting reaction, hence -1/2. 1 𝐾𝐾2 = 𝐾𝐾1−0.5 = (0.0365)−0.5 = 5.23 𝑎𝑎𝑎𝑎𝑚𝑚−2 Q3 What is the equilibrium concentration of Pb2+ ion in a solution of PbX2 that is 0.0420 M in X1ion? PbX2(s) [Pb2+] = Pb2+(aq) + 2X1-(aq) 5.67e-04 M K = 1.00e-06 𝐾𝐾 = [𝑋𝑋 −1 ]2 [𝑃𝑃𝑏𝑏 2+] → [𝑃𝑃𝑏𝑏 2+ ] = 𝐾𝐾 [𝑋𝑋 −1 ]2 = 1𝑥𝑥10−6 = 5.67𝑥𝑥10−4 (0.042)2 Q4 HX is a weak acid that reacts with water according to the following equation: H3O1+(aq) + X1-(aq) HX(aq) + H2O(l) Ka = 2.51e-05 What is the equilibrium concentration of hydronium ion in a solution that is 0.0526 M in HX and 0.253 M in X1- ion? [H3O1+] = 𝑲𝑲𝒂𝒂 = 5.22e-06 M [𝐻𝐻3 𝑂𝑂 +][𝑋𝑋 − ] 𝐾𝐾𝑎𝑎 [𝐻𝐻𝐻𝐻] 2.51𝑥𝑥10−5 (0.0526) → [𝐻𝐻3 𝑂𝑂 +] = = = 5.22𝑥𝑥10−6 [𝐻𝐻𝐻𝐻] [𝑋𝑋 − ] 0.253 Q5 At some temperature, K = 16.2 for the following reaction: H2(g) + CO2(g) H2O(g) + CO(g) The initial partial pressures of H2 and CO2 are each 1.90 atm. What is the final pressure of CO? 1.52 PCO = H2(g) + I C E 1.9 -x 19-x 𝐾𝐾 = 𝑥𝑥 2 = 16.2 → (1.9 − 𝑥𝑥)2 atm CO2(g) 1.9 -x 1.9-x H2O(g) + 0 +x x CO(g) 0 +x x 𝑥𝑥 = 1.52 𝑎𝑎𝑎𝑎𝑎𝑎 What is the final total pressure? Ptotal = 3.80 atm Just add the starting partial pressures because the total P will not change. 2(1.9)= 3.8 atm Q6 An equilibrium mixture contains 0.800 mol HI, 0.270 mol I2, and 0.150 mol H2 in a 1.00-L flask. What is the equilibrium constant for the following reaction? 2HI(g) Kc = 0.0633 H2(g) + I2(g) 𝐾𝐾 = [𝐼𝐼2 ][𝐻𝐻2 ] 0.27(0.15) = = 0.0633 [𝐻𝐻𝐻𝐻]2 (0.8)2 How many moles of I2 must be removed in order to double the number of moles of H2 at equilibrium? 0.367 mol I2 2HI(g) H2(g) + E 0.8 0.27 0.15 I 0.8 0.27-x 0.15 C -0.15 +0.15 +0.15 E 0.65 0.42-x 0.3 𝐾𝐾 = [𝐼𝐼2 ][𝐻𝐻2 ] [𝐻𝐻𝐻𝐻]2 → (0.42−𝑥𝑥)(0.3) (0.65)2 = 0.0633 → I2(g) 𝑥𝑥 = 0.367 mol I2 Hint : If you remove I2, you have new "initial" concentrations. Think about which of these you know, and which final concentrations you know.