tutorial 2 DAS12203

Tutorial 2

1.	(a)	At -15 °C, a natural gas storage tank has volume of 3.25 x 103 L. what is the volume at 31 °C? (3.83 x 103 L)
	(b)	A 50.0 L sample of gas is at 3.00 atm of pressure and a temperature of 298 K. What volume would the gas occupy at STP? (137.4 mL)
	(c)	Tennis balls are filled with nitrogen gas, N2 to increase their “bounce”. Find the pressure inside the tennis balls if it contains 0.33 g of N2 gas at 24 °C and given the volume of the ball is 144 mL. (RAM: N = , R = ) (2 atm)
2.	(a)	For the reaction : NO(g)  +  O2(g)   à   N2O3(g) (i)                 Balance the equation (2NO(g) + 1/2O2(g)  à  N2O3(g)) (ii)  Express the rate in terms of the change in concentration with time for each substance. (iii)             How fast is [O2] decreasing when [NO] is decreasing at a rate of 1.60 x 10-4 mol/L.s? (4 x 10-5 mol/L.s)
	(b)	For the reaction: CF4(g) + H2(g) à CHF3(g) + HF(g), the following data were obtained at 1100 K Exp [CF4(g)] mol/L [H2(g)] (mol/L) Initial rate (mol/L.s) 1 2 3 4 0.1 0.15 0.20 0.20 0.10 0.10 0.20 0.30 45.0 67.5 180 270 (i)                 Write the rate law of the reaction. (x = 1, y = 1) (ii)               Calculate the average value of the rate constant. (4500)
3.	(a)	A layer of copper welded to the bottom of a skillet weighs 125 g. how much heat is needed to raise the temperature of the copper layer from 25 °C to 300 °C? (sCu = 0.387 J/g.C) (13.303 kJ)
	(b)	When 1 mol of KBr(s) decomposes to its elements, 90.29 kJ of heat is absorbed. (i)    Write the balance thermochemical equation for this reaction. (2K(s) + Br2(l)  à  2KBr(s)    ∆H = -90.29 X 2 kJ) (ii)  How much heat is released when 13.5 kg of KBr forms from its elements (reverse reaction from (i)) (RAM: K = , Br = ) (1.02 X 104 kJ )
4.	(a)	Draw Lewis structure for the following molecules, including all formal charges. Also include resonance structure where appropriate. (i)                 CO (ii)               CO2 (iii)             CH4 (iv)             MgCl2 (v)               BeI2
5.	(a)	A 4.50 g gold nugget (specific heat, 0.129 J/g °C) absorbed 276 J of heat. Calculate the final temperature of the gold nugget if the initial T was 25.0 °C (500.5 °C)
	(b)	Given the following thermochemical equation: 3C(s) +2Fe2O3(s) à 4Fe(s) + 3CO2(g)  ∆H = 466 kJ (i)                 Calculate ∆H for the reaction of 2 moles of C. (310.7 kJ0 (ii)               How much heat is required to produce 1000 kg of Fe? (RAM: C= , O=, Fe=,) (2.1 x 106 kJ)
	(c)	Calculate ∆Hf° FeS from the following thermochemical equation: 4FeS(s) + 7O2(g) à 2Fe2O3(s) + 4SO2(g) ∆H = -2435.6 kJ Given: ∆Hf° Fe2O3 = -824.2 kJ/m;, ∆Hf°SO2 = -296.8 kJ/mol (-100 kJ/mol)