CHEM_100/11 2012

ASSIGNMENT #3 Due to November 9

GAS LAWS and Chemical equilibrium

1. {4} The atmospheric pressure today in Antigonish is 740.0 mmHg.

(a) {2} What is the pressure in kPa?

(b) {2} What is the pressure in atm?

Solution:

(a) P_kPa =740.0 mmHg x 101.325kPa/760 mmHg = 98.65855 kPa = 98.66 kPa

(b) P_atm = 740.0 mmHg x 1 atm/760 mmHg = 0.9737 atm

2. {6} Calculate the pressure (in atm) of a sample of O₂ gas (12.0g) in a steel vessel of 15.0 L at 18.0 ^oC.

PV=nRT; P= nRT/V

mol O₂ : 12.0 g x 1mol/32.0g = 0.375 mol

R = 0.0821 L atm/mol K

T= 18 + 273 =291K

P = (0.375 mol x 0.0821 L atm/mol K x 291 K) / 15.0L =0.597 atm

3. {5} A propellant in an aerosol spray has a pressure of 5.00 atm at 16^o C. Instruction on the can includes “do not expose to heat above 60^o C or explosion may result”. What is the pressure inside the can at 61^oC?

Solution :

P₁V₁/n₁T₁ = P₂V₂/n₂T₂ = R (constant) ;

since V and n are constant, P₁/T₁ = P₂/T₂ P₂ = P₁ T₂/T₁

T₁ = 16 +273 = 289 K

T₂ = 61 + 273 = 334 K

P₂ = 5.00 atm x 334 K/ 289 K = 5.8 atm

4. {10} You synthesized a greenish-yellow gaseous compound of chlorine and oxygen and found that its density, d, is 7.71 g/L at 36^oC and 2.88 atm.

(a) {5} Calculate the molar mass, M.

(b) {5} Using molar mass and atomic masses of Cl and O, determine the molecular formula by trial and error.

(a) PV=nRT; n/V= P/RT; nM/V= MP/RT; d= MP/RT; M = dRT/P

M= 7.71 g/L (0.0821 L atm/K mol) (36+273)K/2.88atm = 67.9 g/mol

(b) Cl- 35.45 g, O- 16.00g

One Cl + one O- 51.45 (too low);

2 Cl +1O = 86.9 – too high

1Cl +2O = 67.45 - the formula should be ClO₂.

5. {10} It takes 68.3 s for a sample of N₂ (g) to effuse through a tiny hole. Determine the molecular mass of gas whose effusion time under exactly the same conditions is 85.6s.

(remember that rate and time are inversely proportional)

r₁/r₂ = √ M₂/M₁ = t₂/t₁ ; t₂²/t₁² = M₂/M₁

a) 68.3² s²/85.6²s² = 28 g/mol/M₁ = 0.637 ; M₁ = 28g/mol / 0.637 = 44 g/mol (probably, CO₂)

7. {10} Write the expressions for K_c for the following reactions. In each case indicate whether the rxn is homogeneous or heterogeneous.

(a) N₂ (g) + O₂ (g) à 2NO (g)

homogen

(b) Mo (s) + 2Cl₂ (g) à MoCl₄ (l)

heterogen

(c) 2C₂H₄ (g) + 2H₂O (g) à 2 C₂H₆ (g) + O₂ (g)

homogen

(d) FeO (s) + H₂ (g) à Fe (s) + H₂O (l)

heterogen

(e) 4HBr (aq) + O₂ (g) à H₂O (l) + 2Br₂ (l)

heterogen

8 {10} Determine the value of K_eq for the reaction of sodium and carbon dioxide in water using the data given below

2Na(s) + 2H₂O(l) + 2CO₂(g) D 2NaHCO₃ (s) + H₂ (g)

Na (s) + H₂O (l) D NaOH (aq) + ½ H₂ (g) K_eq = 7.3 x 10³¹

Na₂CO₃ (s) + H₂O (l) D 2NaOH (aq) + CO₂ (g) K_eq = 6.6 x 10^-9

2NaHCO₃ (s) D Na₂CO₃ (s) + H₂O (l) + CO₂ (g) K_eq = 1.1 x 10^-5

Solution:

2Na (s) + 2H₂O (l) D 2NaOH (aq) + H₂ (g) K_eq = (7.3 x 10³¹)²

+

Na₂CO₃ (s) + H₂O (l) + CO₂ (g) D2NaHCO₃ (s) K_eq = (1.1 x 10^-5)^-1

+

2NaOH (aq) + CO₂ (g) D Na₂CO₃ (s) + H₂O (l) K_eq = (6.6 x 10^-9)^-1

____________________________________________________________

2Na(s) + 2H₂O(l) + 2CO₂(g) D 2NaHCO₃ (s) + H₂ (g)

K_eq = (7.3 x 10³¹)² x K_eq = (1.1 x 10^-5)^{-1 x} (6.6 x 10^-9)^-1 = 7.3 x 10⁷⁶

9. {10} At 22 ^oC, K_P = 0.070 for the equilibrium

NH₄HS (s) à NH₃ (g) + H₂S (g)

A sample of solid NH₄HS is placed in a closed vessel and allowed to equilibrate.

(a) {8} Calculate the equilibrium partial pressure (atm) of ammonia, assuming that some solid NH₄HS remains.

b) {2} What is the total pressure in the vessel at equilibrium?

a) K_p = P_NH3 x P_H2S P_NH3 = P_H2S P_NH3 = √K_p

P_NH3 = √ 0.070 = 0.26 atm

b) P_NH3 = P_H2S P_t = P_NH3 + P_H2S

P _t = 0.26 + 0.26 =0.52 atm

10. {10} Consider the following reaction at equilibrium:

2CO₂ (g) à 2CO (g) + O₂ (g) ΔH = -514 kJ

How do the following changes affect the amount of CO(g) ?

(a) increase temperature decrease

(b) more CO₂ added increase

(c) more O₂ added decrease

(c) volume of the vessel is tripled incr.

(d) O₂ is removed incr.

(e) Ar gas is added no effect

(f) total pressure is increased decr.

(g) a catalyst is added no effect

(h) remove some CO₂ (g) decr.

(i) decrease temperature incr.

11 {5} For the equilibrium

N₂ (g) + 3 H₂ (g) D 2NH₃ (g)

the K_p = 4.51 x10^-5 at 450 ^oC.

For the mixture below indicate the direction of this equilibrium reaction (toward product or toward reactants) at the following partial pressures:

105 atm NH₃; 35 atm N₂ ; 495 atm H₂

Q = (105)²/(495)³ x 35 = 2.6 x 10^-6

Q<K_eq therefore rxn will shift to the product.

12. {10} at 100⁰C, K_c = 0.078 for the rxn:

SO₂Cl₂ (g) D SO₂ (g) + Cl₂ (g)

At the equilibrium, the concentrations of SO₂Cl₂ and SO₂ are 0.108 M and 0.052 M, respectively. What is the partial pressure of Cl₂ at equilibrium? (hint: modify the ideal gas equation using the fact that n/V=M )

Solution:

K_c = [SO₂] [Cl₂]/[SO₂Cl₂] à [Cl₂] = K_c [[SO₂Cl₂]/ [SO₂] = 0.078x0.108/0.052= 0.16M

PV = nRT; PV/V = (n/V) RT; P= MRT

P = 0.16 mol/L x 0.0821 L atm mol^-1K^-1 373 K = 4.9 atm