Chemistry Second Exam 2020

General Chem Second Exam Time: 60 min. Student's Name: .......... Section No. : .......... Physical constants and useful relations: 1 atm = 101.3 kPa = 1.013 x 10^5 Pa = 760. Torr; Planck's constant = 6.63x10^-34 J.s, Speed of light = 3.00 x 10^8 m/s; E_n = -(2.18 x10^-18/n^2) J; E = hc/λ R = 0.08206 L. atm/mol. K = 8.314 J/mol.K 1L. atm = 101.3 J; λ = h/mu Av. No. = 6.022 x 10^23 mol^-1; PV = n RT; [P + a (n/V)^2 ] (V - n b) = nRT u_rms = (3RT/M)^1/2; ΔE = q + w, ΔH = ΔE + P x ΔV, w = -P x ΔV ANSWER SHEET a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e a b c d e Which of the following statements concerning gases is correct? a) All gases behave ideally at high P and/or low T. b) No gases behave ideally at low P and/or high T. c) No gases behave ideally at high P and/or high T. d) All gases behave ideally at low P and/or high T. e) Both van der Waals constants (a & b) are the same for all gases. (handwritten notes): Real = low T, high P Ideal = low P, high T Which of the following statements concerning ideal gases is incorrect (not correct)? a) At constant n and T, P1 V1 = P2 V2. b) The average molecular speed is higher for H2 gas than for N2 gas at same T. c) The average kinetic energy is higher for H2 gas than for N2 gas at same T. d) At constant n, P1/T1 = P2/T2. e) At constant n and V, P1/T1 = P2/T2. According to Kinetic Molecular Theory of gases, the root-mean square speed (u_rms) of N2 gas (M = 28.0 g/mol ) at 25°C is equal to a) 411 m/s b) 515 m/s c) 610. m/s d) 682 m/s e) 742 m/s u_rms = √(3RT / M) = √(3 * 298 * 8.314 / 28 x 10^-3) Given that the density for an ideal gas (d = 1.801 g/L) at 1.00 atm and 25°C, the molar mass (M in g/mol ) of the gas is equal to a) 44.0 g/mol b) 30.0 g/mol c) 610. g/mol d) 58.0 g/mol e) 72.0 g/mol M= dRT / P = (1.801 * 0.082 * 298) /1 M= dRT / P The nitrogen (N2) gas obtained from the decomposition of sodium azide (NaN3) according to the chemical reaction: 2 NaN3(s) -> 2 Na(s) + 3 N2(g) was collected over liquid water at a total pressure of 724 torr and 25°C where the vapor pressure of water was 24.0 torr. If the volume of the N2 gas was 10.0 L, then the mass of N2 gas is equal to.... (Molar mass of N2 = 28.0 g/mole ) m = PVM / RT a) 5.27 g b) 15.8. g c) 10.5 g d) 21.1 g e) 38.7 g m= (724-24) / 760 x 10 / 0.082 x 298 x 28 = 10.5 P = nRT / V m = nRT / v (handwritten calculations and notes) P = nRT / M 0.95 atm P = nRT / V 6. Given that 4.00 g of CH4 gas (M = 16.0 g/mol ) and 22.0 g of C3H8 gas (M = 44.0 g/mol ) were placed in a 25.0 L container at 25°C, then the total pressure (P in kPa) of the gas mixture would be equal to a) 92.9 kPa b) 61.9 kPa c) 74.3 kPa d) 53.1 kPa e) 40.8 kPa P = nRT / V = ( 4/16 + 22/44) 0.082 x 298 /25 = 1.7 atm = 1.7 x 101.3 kPa / 1 atm = 170 / 25 = 74.3 The combustion (oxidation) of propane (M = 44.0 g/mole ) is described by the balanced equation C3H8(g) + 5 O2(g) --> 3 CO2(g) + 4 H2O(l) ΔH = -2220 kJ Calculate the mass of propane (in gram) must be burned to produce 175.5 kJ of heat. a- 6.96 b- 13.9 c- 20.9 d- 3.48 e- 4.40 -2220 / x = 175.5 / 44 x = 3.48 Given the following data: 2 C6H6(l) + 15 O2(g) --> 12 CO2(g) + 6 H2O(l) ΔH°f ( kJ/mol) +49 -393.51 -285.83 Calculate the standard enthalpy of combustion of benzene in (kJ/mole benzene) a) -3135.5 b) 6535.2 c) -6270.9 d) -3267.6 e) -6535.2 product - reactant = (12 x -393.51 + 6 x -285.83) - ( 2 x 49) = -4722.12 + -1714.98 = -6437 + 98 = -6340 A gas is allowed to expand, at constant temperature, from a volume of 1.0 L to 10.1 L against an external pressure of 0.50 atm. If the gas absorbs 250 J of heat from the surroundings, what is the value of q, w, and ΔE? q w ΔE a) 250 J -4.55 J 245 J b) -250 J -460 J -710 J c) 250 J 460 J 710 J d) -250 J 460 J 210 J e) 250 J -460 J -210 J w = - PΔV = -.5 ( 10.1 -1) atm L x 101.3 = -460J ΔE = q + w = 250 - 460 = -210 A 100.0 ml of 0.200 M aqueous hydrochloric acid, is added to 100.0 ml of 0.200 M aqueous ammonia (NH3) in a constant pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 25.00 °C. The final temperature after mixing is 26.20 °C. Assuming the density of the solution = 1.00 g/ml and its specific heat = 4.18 J/g.°C, calculate ΔH per mole of the reaction: HCl(aq) + NH3(aq) --> NH4Cl(aq) a) -58.6 kJ b) -54.3 kJ c) -50.2kJ d) -42.2 kJ e) -46.0 kJ q = msΔT = (200) x 4.18 x ( 1.20) = -1000 ΔH = q = 1000 / 0.02 = -50000