Hw 1 ch 302 - obennoskey(blo242 Homework#1 Holcombe(52460...

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obennoskey (blo242) – Homework #1 – Holcombe – (52460) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What factor determines whether a reaction will be exothermic or endothermic? 1. the relative concentrations of reactants and products in solution 2. the activation energy of the reaction 3. the speed of the reaction 4. the heat properties of the products 5. the difference in energy between reactants and products correct Explanation: Δ H is the energy difference between the products and reactants; Δ H for exothermic reactions is negative, and Δ H for endother- mic reactions is positive. 002 10.0 points What is true about the first law of thermody- namics? 1. Δ E univ > 0 2. Δ E univ = 0 correct 3. Δ E sys > 0 4. Δ E univ < 0 5. Δ E sys < 0 6. Δ E sys = 0 Explanation: 003 10.0 points In the manufacture of nitric acid by the oxi- dation of ammonia, the first product is nitric oxide. The nitric oxide is then oxidized to nitrogen dioxide: 2 NO(g) + O 2 (g) -→ 2 NO 2 (g) Calculate the standard reaction enthalpy for the reaction above (as written) using the following data: N 2 (g) + O 2 (g) -→ 2 NO(g) Δ H = 180 . 5 kJ N 2 (g) + 2 O 2 (g) -→ 2 NO 2 (g) Δ H = 66 . 4 kJ 1. - 975 . 0 kJ/mol rxn 2. - 114 . 1 kJ/mol rxn correct 3. - 690 . 72 kJ/mol rxn 4. - 520 . 2 kJ/mol rxn 5. - 128 . 2 kJ/mol rxn 6. - 252 . 4 kJ/mol rxn 7. - 100 . 3 kJ/mol rxn Explanation: Using Hess’ Law and the given standard re- action enthalpies, the first reaction is reversed and added to the second: 2 NO(g) -→ N 2 (g) + O 2 (g) Δ H = - 180 . 5 kJ N 2 (g) + 2 O 2 (g) -→ 2 NO 2 (g) Δ H = 66 . 4 kJ 2 NO(g) + O 2 (g) -→ 2 NO 2 (g) Δ H = - 114 . 1 kJ 004 10.0 points A piece of a newly synthesized material of mass 25.0 g at 80.0 C is placed in a calorime- ter containing 100.0 g of water at 20.0 C. If the final temperature of the system is 24.0 C, what is the specific heat capacity of this ma- terial? 1. 4 . 76 J · g 1 · ( C) 1 2. 7 . 46 J · g 1 · ( C) 1
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obennoskey (blo242) – Homework #1 – Holcombe – (52460) 2 3. 0 . 84 J · g 1 · ( C) 1 4. 1 . 19 J · g 1 · ( C) 1 correct 5. 0 . 30 J · g 1 · ( C) 1 Explanation: Δ T m = (24 C - 80 C) = - 56 C Δ T w = (24 C - 20 C) = 4 C C s w = 4 . 184 J · g 1 · C 1 m w = 100 g m m = 25 g Heat energy will flow out of the hot metal in to the cool water. The final temperature of BOTH items will be 24.0 C. We can set up two expressions for heat flow using heat capacity and set them equal and opposite to each other in sign: - m m C s , m Δ T m = m w C s , w Δ T w C s , m = - m w C s , w Δ T w m m Δ T m C s , m = - (100 g)(4 C) (25 g)( - 56 C) × (4 . 184 J · g 1 · C 1 ) = 1 . 19543 J · g 1 · C 1 005 10.0 points What mass of ethanol (C 2 H 5 OH( )) must be burned to supply 500 kJ of heat? The stan- dard enthalpy of combustion of ethanol at 298 K is - 1368 kJ · mol 1 .
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