HW 4 Sol'n - dayton(mrd772 – Homework 4 – sutcliffe...

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Unformatted text preview: dayton (mrd772) – Homework 4 – sutcliffe – (51060) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. There are more complex equilibrium cal- culations here. You may have to use the quadratic formula to solve some. Make sure you know the formula! This HW covers the rest of Chapter 6. 001 10.0 points At a certain temperature, the equilibrium constant K c is 0.154 for the reaction 2 SO 2 (g) + O 2 (g) ⇀ ↽ 2 SO 3 (g) What concentration of SO 3 would be in equilibrium with 0.250 moles of SO 2 and 0 . 583 moles of O 2 in a 1.00 liter container at this temperature? Note: These latter moles are the equilibrium values. Correct answer: 0 . 0749091 M. Explanation: K c = 0.154 n SO 2 = 0.25 mol n O 2 = 0 . 583 mol V container = 1.0 L 2 SO 2 (g) + O 2 (g) ⇀ ↽ 2 SO 3 (g) [SO 2 ] = . 250 mol 1 . 00 L = 0 . 250 M [O 2 ] = . 583 mol 1 . 00 L = 0 . 583 M K c = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] = 0 . 154 [SO 3 ] = radicalBig K c [SO 2 ] 2 [O 2 ] = radicalBig . 154 × (0 . 250) 2 × . 583 = 0 . 0749091 M 002 10.0 points Note: Use K c . For the reaction 2 NOCl(g) → 2 NO(g) + Cl 2 (g) , if only [NOCl(g)] = 2.8 M initially, then at equilibrium [NO(g)]= 1.2 M. What is the equilibrium concentration of NOCl(g)? As- sume that only reactants exist initially. 1. 3.2 M 2. 0.8 M 3. 1.6 M correct 4. 2.2 M 5. 1.2 M Explanation: 003 10.0 points A 25 g sample of ammonium carbamate (NH 4 (NH 2 CO 2 )), was placed in an evacuated . 25 L flask and kept at 25 ◦ C. At equilibrium, 17 . 4 mg of CO 2 was present. NH 4 (NH 2 CO 2 )(s) ⇀ ↽ 2 NH 3 (g) + CO 2 (g) . What is the value of K c for the decomposi- tion of ammonium carbamate into ammonia and carbon dioxide? Correct answer: 1 . 5821 × 10 − 8 . Explanation: 25 g NH 4 (NH 2 CO 2 ) 78 . 07 g / mol NH 4 (NH 2 CO 2 ) = 0 . 320225 moles NH 4 (NH 2 CO 2 ) At equlibrium, . 0174 g CO 2 44 . 01 g / mol CO 2 = 0 . 000395365 moles CO 2 From the balanced reaction, 2 mol NH 3 are formed per mol of CO 2 formed, so n NH 3 = 2(0 . 000395365 mol) = 0 . 000790729 mol K c = [NH 3 ] 2 [CO 2 ] = parenleftbigg . 000790729 . 25 L parenrightbigg 2 parenleftbigg . 000395365 . 25 L parenrightbigg = 1 . 5821 × 10 − 8 dayton (mrd772) – Homework 4 – sutcliffe – (51060) 2 004 10.0 points At T = 600 ◦ C, K c = 144 for the gas-phase reaction A + B ⇀ ↽ C + D Starting with 1 . 35 moles each of A and B in a 5.00 liter container, what will be the equilib- rium concentration of C at this temperature? Correct answer: 0 . 249231 M. Explanation: [A] = 1 . 35 mol 5 L = 0 . 27 M T = 600 ◦ C [B] = 1 . 35 mol 5 L = 0 . 27 M K c = 144 A + B ⇀ ↽ C + D ini, M . 27 . 27 Δ, M- x- x x x eq, M . 27- x . 27- x x x [C][D] [A][B] = 144 x 2 (0 . 27- x ) 2 = 144 x . 27- x = 12 x = 3 . 24- 12 x x = [C] = 0 . 249231 M 005 10.0 points Suppose the reaction H 2 (g) + I 2 (g) ⇀ ↽ 2 HI(g) has an equilibrium constant K c = 49 and the initial concentrations of H...
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HW 4 Sol'n - dayton(mrd772 – Homework 4 – sutcliffe...

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