CH-Exam 1-Exam 2 - Version 433 – Exam 1 – Sutcliffe –...

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Unformatted text preview: Version 433 – Exam 1 – Sutcliffe – (52410) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Calorimetry can be used to find heats of com- bustion. Suppose a calorimeter helps some- one to determine that 4032 kJ of heat is re- leased during the combustion of 88.0 grams of propane, C 3 H 8 (44.0 g/mol). What would be the amount of heat released for the combus- tion of 0.500 moles of propane? 1. 4032 kJ 2. 1008 kJ correct 3. 504 kJ 4. 2016 kJ 5. 8064 kJ Explanation: q = 4032 kJ m = 88.0 g n = 0.500 mol MW = 44.0 g/mol If we know how much heat is evolved from combusting 88 g of propane, then we can use the molecular weight of propane to calculate how much heat is evolved from the combus- tion of 0.5 moles: 4032 kJ 88 . 0 g C 3 H 8 44 . 0 g C 3 H 8 1 mol C 3 H 8 × . 5 mol C 3 H 8 = 1008 kJ 002 10.0 points How much energy is required to complete the transition 50 g H 2 O( ) at 40 ◦ C → 50 g H 2 O(g) at 100 ◦ C ? 1. 35.4 kcal 2. 0.3 kcal 3. 7.0 kcal 4. 30 kcal correct 5. 27 kcal Explanation: 50 g H 2 O( ) 40 ◦ C step 1-→ 50 g H 2 O( ) 100 ◦ C step 2-→ 50 g H 2 O(g) 100 ◦ C Step 1: 1 cal g · ◦ C · (50 g) · (100- 40) ◦ C = 3000 cal Step 2: 540 cal g · (50 g) = 27000 cal Total = 27000 cal + 3000 cal = 30000 cal = 30 kcal 003 10.0 points Calculate the work for the reaction 16 CO 2 (g) + 18 H 2 O(g) → 2 C 8 H 18 ( ) + 25 O 2 (g) at 298 K. 1.- 22 . 30 kJ/mol 2.- 17 . 34 kJ/mol 3. +9 . 00 kJ/mol 4. +22 . 30 kJ/mol correct 5. +17 . 34 kJ/mol 6.- 9 . 00 kJ/mol Explanation: The number of moles is Δ n = 25 mol- (16 mol + 18 mol) =- 9 mol Thus the work is w =- Δ nRT =- (- 9 mol) (8 . 314 J / mol · K) (298 K) = 22298 . 1 J / mol = 22 . 2981 kJ / mol 004 10.0 points Version 433 – Exam 1 – Sutcliffe – (52410) 2 Which one of the following reactions has a positive entropy change? 1. 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g) 2. H 2 O(g) → H 2 O( ) 3. BF 3 (g) + NH 3 (g) → F 3 BNH 3 (s) 4. 2 NH 4 NO 3 (s) → 2 N 2 (g) + 4 H 2 O(g) + O 2 (g) correct 5. N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) Explanation: Processes that result in an increase in the number of moles of gaseous substances have ΔS sys > 0. Only this choice demonstrates an increase in the number of moles of gaseous substances (from 0 moles of gaseous reactants to 7 moles of gaseous products). 005 10.0 points Consider the reaction 2 Fe 2 O 3 (s) + 3 C(s) → 4 Fe(s) + 3 CO 2 (g) , Δ H ◦ = 462 kJ , Δ S ◦ = 558 J · K- 1 . Calcu- late the equilibrium constant for this reaction at 525 ◦ C. 1. 5 . 20 × 10- 7 2. 1 . 9 × 10 6 3. 3 . 04 × 10- 3 4. 8 . 07 × 10- 2 correct 5. 2 . 18 × 10- 2 Explanation: 006 10.0 points Determine the enthalpy change of reaction at 25 ◦ C for 4 HNO 3 ( ) + 5 N 2 H 4 ( ) → 7 N 2 (g) + 12 H 2 O(g) Δ H f for HNO 3 is- 174 . 1 kJ/mol; Δ H f for H 2 O is- 241 . 8 kJ/mol; Δ H f for N 2 H 4 is +50 . 63 kJ/mol....
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This note was uploaded on 02/13/2012 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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CH-Exam 1-Exam 2 - Version 433 – Exam 1 – Sutcliffe –...

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