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Test1FormASolutions - Chemistry 1036 Test 1 THIS IS FORM...

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Chemistry 1036 Test 1 February 7, 2008 THIS IS FORM A [FORM B #s in ( )] Name __________________________ Choose the single best answer to each question. 1.(4) If 600 J of heat is added to a system and the system does 450 J of work on the surroundings, what is the change in internal energy of the system? 1. 1050 J 2. 150 J 3. –1050 J 4. –150 J Answer: 2. 150 J E = q + w q = +600 J; heat is added to the system; positive value w = -450 J; the system does work; negative value E = 600 J + (-450 J) E = 150 J 2.(8) One mole of a gas expands from a volume of 2.50 L to a volume of 10.00 L against a constant external pressure of 0.975 atm. How much work is done on the surroundings? (1 L atm = 101.3 J). 1. 741 J 2. 988 J 3. 611 J 4. 760. J Answer: 1. 741 J w = -P V P = 0.975 atm V 2 = 10.00 L V 1 = 2.50 L V = V 2 – V 1 = 10.00 L – 2.50 L = 7.50 L w – 1(0.975 atm)(7.50 L) w = -7.31 L atm 7.31 L atm 101.3 J x 1 L atm = -741 J 741 J of work is done by the system on the surroundings 3.(6) Calculate H o for the reaction: 2 N 2 (g) + 5 O 2 (g) 2 N 2 O 5 (g)

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given the following thermochemical equations: N 2 (g) + 3 O 2 (g) + H 2 (g) 2 HNO 3 (aq) H o = -414.0 kJ N 2 O 5 (s) + H 2 O( A ) 2 HNO 3 (aq) H o = -86.0 kJ 2 H 2 (g) + O 2 (g) 2 H 2 O( A ) H o = -571.6 kJ 1. –84.4 kJ 2. –1571 kJ 3. 243.6 kJ 4. -121.8 kJ Answer: 1. –84.4 kJ The reactant N 2 is also the reactant in the first reaction but the coefficient must be 2 instead of 1. Multiply Reaction 1 by 2: 2 N 2 (g) + 6 O 2 (g) + 2 H 2 (g) 4 HNO 3 (aq) H o = 2(-414.0 kJ) = -828 kJ Reverse the second reaction and multiply by 2 to obtain 2 moles of N 2 O 5 as a product. Also multiply the value of H by 2 and change the sign from negative to positive since the reaction was reversed: 4 HNO 3 (aq) 2 N 2 O 5 (s) + 2 H 2 O( A ) H o = 2(+86.0 kJ) = +172 kJ There are 2 moles of H 2 O on the right side of equation 2. To eliminate the H 2 O, reverse Reaction 3 so that H 2 O is a reactant and change the sign of the enthalpy: 2 H 2 O( A ) 2 H 2 (g) + O 2 (g) H o = +571.6 kJ Add the three reaction is bold: 2 N 2 (g) + 6 O 2 (g) + 2 H 2 (g) 4 HNO 3 (aq) H o = -828 kJ 4 HNO 3 (aq) 2 N 2 O 5 (s) + 2 H 2 O( A ) H o = +172 kJ 2 H 2 O( A ) 2 H 2 (g) + O 2 (g) H o = +571.6 kJ 2N 2 + 6 O 2 + 2 H 2 + 4 HNO 3 + 2 H 2 O 4 HNO 3 + 2 N 2 O 5 + 2 H 2 O + 2 H 2 + O 2 5 2 N 2 (g) + 5 O 2 (g) 2 N 2 O 5 (g) H o = -828 kJ + 172 + 571.6 kJ = -84.4 kJ (1 mole of O 2 on the left cancels the 1 mole of O 2 on the right, leaving you with 5 moles of O 2 on the left. The 2 moles of H 2 on the left cancels the 2 moles of H 2 on the right, the 4 moles of HNO 3 cancels the 4 moles of HNO 3 on the right and the 2 moles of H 2 O on the left cancels the 2 moles of H 2 O on the right.) 4.(3) In which of the following reactions is work done on the system? 1. CaC 2 (s) + 2 H 2 O( A ) Ca(OH) 2 (aq) + C 2 H 2 (g) 2. P 4 O 10 (s) + 6 H 2 O( A ) 4 H 3 PO 4 (s)
3. P 4 (s) + 6 Cl 2 (g) 4 PCl 3 (g) 4. C(s) + CO 2 (g) 2 CO(g) Answer: 3. P 4 (s) + 6 Cl 2 (g) 4 PCl 3 (g) 1. CaC 2 (s) + 2 H 2 O( A ) Ca(OH) 2 (aq) + C 2 H 2 (g) 0 moles of gas 1 mole of gas The system is expanding; there are 0 moles of reactant gas and 1 mole of product gas. The volume of the system increases since the number of moles of gas increases; the system is doing work on the surroundings as it expands.

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