1 hi 0 233 233 12 2 hi 0 233 233 12 1 2 3 hi

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1. [HI] = (0 . 233 · 0 . 233 · 0 . 12) 2. [HI] = (0 . 233 · 0 . 233 · 0 . 12) 1 / 2 3. [HI] = parenleftbigg 0 . 233 0 . 12 parenrightbigg 1 / 2 4. [HI] = parenleftbigg 0 . 233 · 0 . 233 0 . 12 parenrightbigg 5. [HI] = parenleftbigg 0 . 233 · 0 . 233 0 . 12 parenrightbigg 1 / 2 correct Explanation: Since we started from pure HI, the equilib- rium concentrations of hydrogen and iodine gas must be equal. 021 3.6points K c = 58 at some temperature for the reaction H 2 (g) + I 2 (g) 2 HI(g) . If 30 . 4 mol of HI are introduced into a 10.0 liter vessel, how many moles of I 2 are present at equilibrium? 1. 3 . 79377 mol 2. 4 . 74221 mol 3. 3 . 16147 mol correct 4. 1 . 58074 mol 5. 6 . 32294 mol 6. 31 . 6147 mol Explanation: K c = 58 V vessel = 10.0 L n HI = 30 . 4 [HI] ini = 30 . 4 mol 10 L = 3 . 04 M H 2 (g) + I 2 (g) 2 HI (g) ini, M 0 0 3 . 04 Δ, M x x - 2 x eq, M x x 3 . 04 - 2 x K c = [HI] 2 [H 2 ] [I 2 ] = 58 (3 . 04 - 2 x ) 2 x 2 = 58 3 . 04 - 2 x x = 58 3 . 04 - 2 x = 58 x x = 0 . 316147 n I 2 = (10 . 0 L) [I 2 ] = (10 . 0 L) parenleftbigg 0 . 316147 mol L parenrightbigg = 3 . 16147 mol 022 3.6points Consider the following reaction: PCl 5 (g) PCl 3 (g) + Cl 2 (g) . At equilibrium, [PCl 5 ] = 2.00 M and [PCl 3 ] = [Cl 2 ] = 1.00 M. If suddenly 1.00 M PCl 5 (g), PCl 3 (g), and Cl 2 (g) are each added, calculate the equilibrium concentration of PCl 3 (g). 1. 1.95 M
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casey (rmc2555) – Homework 5 – holcombe – (51395) 7 2. 1.43 M 3. 2.32 M 4. 1.35 M correct 5. 3.35 M 6. 2.18 M Explanation: K = (1)(1) 2 = 0 . 5 after the addition each concentration in- creases by 1 M and then the reaction must shift to the left to reestablish equilibrium. K = (2 - x ) 2 (3 + x ) = 0 . 5 which converts to 0 = x 2 - 4 . 5 x + 2 . 5 The quadratic formula will yield two an- swers for x . x = 0 . 64929 or 3 . 85078 The 3.85078 answer is not physically possible and therefore x = 0 . 64929. [PCl 3 ] = 2 - x = 1 . 35078 M 023 3.6points Suppose the reaction H 2 (g) + I 2 (g) 2 HI(g) has an equilibrium constant K c = 49 and the initial concentration of H 2 and I 2 is 0.5 M and HI is 0.0 M. Which of the following is the correct value for the final concentration of HI(g)? 1. 0.250 M 2. 0.219 M 3. 0.599 M 4. 0.778 M correct 5. 0.389 M Explanation: K c = 49 [H 2 ] ini = 0 . 5 M [I 2 ] ini = 0 . 5 M [HI] ini = 0 M H 2 (g) + I 2 (g) 2 HI(g) Ini, M 0.5 0.5 0 Δ, M - x - x +2 x Equil, M 0 . 5 - x 0 . 5 - x 2 x K c = [HI] 2 [H 2 ] [I 2 ] 49 = (2 x ) 2 (0 . 5 - x ) 2 7 = 2 x 0 . 5 - x 7(0 . 5 - x ) = 2 x 3 . 5 - 7 x = 2 x 3 . 5 = 9 x x = 3 . 5 9 = 0 . 389 M Looking back at our equilibrium values, we see that the final concentration of HI is equal to 2 x , so 2(0.389) = 0.778 M. 024 3.6points Suppose the reaction A B has an equilibrium constant of 1.0 and the ini- tial concentrations of A and B are 0.5 M and 0.0 M, respectively. Which of the following is the correct value for the final concentration of A? 1. None of these is correct. 2. 0.250 M correct 3. 1.00 M 4. 1.50 M 5. 0.500 M
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casey (rmc2555) – Homework 5 – holcombe – (51395) 8 Explanation: K = 1 . 0 [A] ini = 0 . 5 M [B] ini = 0 M A B ini, M 0.5 0.0 Δ, M - x x eq, M 0 . 5 - x x K = [B] [A] = 1 . 0 x 0 . 5 - x = 1 . 0 x = 0 . 25 M [A] = 0 . 5 - x = 0 . 25 M 025 3.6points At 25 C, K = 6 . 9 × 10 5 for the reaction N 2 (g) + 3 H 2 (g) 2 NH 3 (g) . Calculate K c at 25 C for this reaction. 1. 4 . 1 × 10 8 correct 2. 6 . 8 × 10 5 3. 1 . 1 × 10 3 4. 1 . 7 × 10 7 5. 2 . 8 × 10 4 Explanation: K is the same as K p in this problem. K p = K c ( RT ) Δ n You must use R = 0 . 08206 L atm/mol K for this problem. The value for Δ n is -2. So solving for K c gives the following equation: K c = K p ( RT ) Δ n K c = 6 . 9 × 10 5 (24 . 47) 2 K c = 4 . 13 × 10 8 026 3.6points Calculate the value of K at 700 K for the reaction H 2 (g) + I 2 (g) 2 HI(g) if K c = 54 at the same temperature.
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