Exam1SW05 - C h e m 5, 9 Section Exam 1 Solutions Winter,...

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Chem 5, 9 Section Winter, 2005 Exam 1 Solutions January 27, 1 1. (10 + 5 points) In preparation for a study of the equilibrium constant for the reaction 2 NO 2 ( g ) N 2 O 4 ( g ) you take an 8.6 g frozen sample of pure N 2 O 4 ( s ) and place it in an otherwise evacuated container of volume 2.3 L and allow it to warm up to room temperature, 300 K, where everything in the container will be a gas. (a) If the equilibrium constant for this reaction, K P , was infinite , what would the pressure be when all the sample vaporized? For the reaction as written above (and if you worked the problem with the reverse reaction but otherwise drew the correct conclusions, you weren’t penalized), under this assumption for K P , we can write K P = P N 2 O 4 P NO 2 2 = which is true if P NO 2 is zero, i.e., all the gas is N 2 O 4 . Thus, the number of moles of gas, n , is equal to the number of moles of N 2 O 4 in the 8.6 g sample: n = m M = 8.6 g 92.011 g mol –1 = 0.0935 mol Thus, the pressure is simply P = nRT V = 0.0935 mol 0.08206 L atm mol –1 K –1 300 K 2.3 L = 1.00 atm (b) In fact, of course, K P is finite. How will the actual final pressure compare to the answer you gave in part (a)? (circle one) greater the same smaller Briefly justify your answer. If K P is finite, then at equilibrium, some of the N 2 O 4 will have dissociated into NO 2 . But every one N 2 O 4 that dissociates produces two NO 2 molecules in the gas. Thus, n must increase, and if T and V are constant, a larger n means a larger P . 2. (10 + 6 + 4 points) A 0.578 g sample of pure tin (Sn) is reacted with excess fluorine gas (F 2 ) until the resultant solid product has a constant mass of 0.948 g. (a) What is the empirical formula of the compound formed? We will need the molar masses of Sn and F: 118.71 g mol –1 and 18.998 g mol –1 , respectively. The mass of Sn tells us the number of moles of Sn: n Sn = m Sn M Sn = 0.578 g 118.71 g mol = 4.87 × 10 –3 mol The mass of F in the product is the difference between the product mass and the original Sn sample mass, and thus the number of moles of F in the product is n F = m F M F = 0.948 g g 18.998 g mol = 1.95 × 10 –2 mol Thus, the empirical formula, which is simply the ratio of the number of F atoms to the number of Sn atoms, is
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Chem 5, 9 Section Winter, 2005 Exam 1 Solutions January 27, 2 n F n Sn = 1.95 × 10 –2 mol 4.87 × 10 –3 mol = 4.00 and the empirical formula is SnF 4 . (b) Like many metal fluorides, this one is fairly volatile. If the solid is volatilized into a 1.5 L container at 500 K, a pressure of 0.133 atm is measured. What does this tell you about the molecular formula of the compound? If we find the number of moles of gas from these data, we can find the molecular formula: n = PV RT = 0.133 atm 1.5 L 0.08206 L atm mol –1 K –1 500 K = 4.87 × 10 –3 mol This is the same number of moles as there are moles of Sn atoms in the compound, and thus each molecule has only one Sn atom.
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This note was uploaded on 05/01/2008 for the course CHEM 5/6 taught by Professor Gribble during the Spring '08 term at Dartmouth.

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Exam1SW05 - C h e m 5, 9 Section Exam 1 Solutions Winter,...

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