Practice+Quiz+7

Practice Quiz 8, New Material Question 1 4 pts Label the following statements about equilibrium as true or false. Adding a catalyst will increase the amount of products. True False The concentrations of reactants and products are constant. True False The rate of the forward reaction equals the rate of the reverse reaction. True False The value of the equilibrium constant will change if the temperature changes True False Question 2 4 pts The following reaction is at equilibrium. NH 4 Cl (s) ⇌ NH 3 (g) + HCl (g) What is the effect of the following changes on the position of the equilibrium? Circle one choice for each part. a) NH 3 (g) is removed shifts to make more products shifts to make more reactants no shift b) HCl (g) is added shifts to make more products shifts to make more reactants no shift c) NH 4 Cl (s) is added. shifts to make more products shifts to make more reactants no shift d) NH 3 (g) is added shifts to make more products shifts to make more reactants no shift Question 3 2 pts Consider the following reaction and equilibrium constant: 2 PH 3 (g) ⇌ P 2 (g) + 3 H 2 (g) K p = 398 at 873 K The initial partial pressures are: Partial pressure PH 3 = 0.0260 atm, Partial pressure P 2 = 0.871 atm Partial Pressure H 2 = 0.517 atm Determine the direction of the reaction to attain equilibrium. You must show your work. Circle one: Shifts toward Products Shifts toward reactants no shift 7 O O energy z O L reaction O progress A 8 B 10 . e given equations : 2 G = C M &T [0 0260] ⊥ G = CH ° TOSO -G ° = -RTInkC Qp = 178 ⊥ G = LG +RTInQ Kp < Op T Shift to products

Question 4 1 pts Methanol can be produced from carbon monoxide and hydrogen with suitable catalysts: CO( g ) + 2 H2( g ) → CH3OH( l ) at 25°C Δ H ° = -128.1 kJ and Δ S ° = -332 J/K Find ∆G° at 25°C. Question 5 1 pt ∆G° for a reaction is -37.2 kJ at 25°C. What is K c for this reaction at the same temperature? Question 6 1 pt Kc for a reaction at 25°C = 0.015. Calculate ∆G° for this reaction at the same temperature. Question 7 1 pt A mixture of 0.500 mole of carbon monoxide and 0.400 mole of bromine were placed into a rigid 1.00-L container and the system was allowed to come to equilibrium. The equilibrium concentration of COBr 2 was 0.233 M . What is the value of K c for this reaction? "# ( $ )+ %& 2( $ ) ⇌"#%& 2( $ )CO(g)+Br2(g) ⇔ COBr2(g) Show your work. -S ° = 0 332k)/K T = 298K LG = WH ° -TOS - G ° = 128 /k - 29870 332k)/k) ↓ G ° = 29 164 /mul -G : RTInKc 37 2k) = 0 008314(298k)InKc 37 2k) = 2 478/nk (nkc = 15 01 K = 3301 S kJ/mU) Kc = 3302)/mo) .. 3 x100k/mol T = 298K OG = -RTInke K = 0 015 -G ° = 8 3/4(298)In 0 015 UG ° = 10405)/mol or C 10 4/k)/mol mus I 0 . 5 0 4 O + - - C 1 x / - 04 - x 0 233 & 0 . 5 - 4 ↑ x = 0 . 233 CoBris [CO] = 0 . 267 PosBre] Colb [Brz) = 0 167