CHEM_100 2011

ASSIGNMENT #4 Due to November 19

Chemical equilibrium (chapter 15 and 16)

SOLUTIONS

1. {10} The 0.42 moles of CO and 0.42 mol of H₂ were placed in an empty 1.00-L container and allowed to reach equilibrium described by the equation:

CO (g) + 2H₂ (g) D CH₃OH (g)

At the equilibrium there were 0.29 mol CO remaining. What is K_eq ?

CO (g) + 2H₂ (g) D CH₃OH (g)

In 0.42 0.42 0

Ch -0.13 -0.13x2 0.13

Eq 0.29 0.16 0.13

K_eq = 0.13/(0.16² x 0.29) = 17.5

2. {15} For the reaction I₂ (g) + Br₂ (g) D 2 IBr (g) , K_c = 280.0 at 150.0⁰ C

1.00 mol of IBr in a 2.00-L flask is allowed to reach equilibrium at 150⁰C. What are the equilibrium concentrations of IBr, Br₂, and I₂ ?

Solution:

[IBr] = 1.00 mol/2.00L = 0.500M

I₂ (g) + Br₂ (g) D 2 IBr (g)

in 0 0 0.5

ch x x -2x

eq x x 0.500-2x

K = (0.500-2x)²/x² = 280 0.500-2x/x = √280 = 16.733

0.500-2x = 16.733x; 18.733x = 0.500; x =0.0267 = [Br₂] = [I₂]

[IBr] = 0.500 -2x0.0267 = 0.447

3. {15} A mixture of 1.374 g H₂ and 70.31 g of Br₂ is heated in a 2-L vessel at 700 K. The reaction runs as follows:

H₂ (g) + Br₂ (g) D 2 HBr (g)

At equilibrium the vessel contains 0.566g of H₂.

a) {10} Calculate the equilibrium concentrations of H₂, Br₂, and HBr.

b) Calculate K_c.

Solution:

[H₂]_initial = 1.374g x 1mol /2.0159g x 1/2.00L = 0.341M

[Br₂]_in = 70.31g x 1mol/159.81g x 1/2.00L = 0.220 M

[H₂]_eq = 0.566g x 1mol/2.0159g x 1/2.00L = 0.140M

H₂ (g) + Br₂ (g) D 2 HBr (g)

in 0.341 0.220 0

ch -0.201 -0.201 0.402

eq 0.140 0.019 0.402

K_c = 0.402²/(0.140x 0.019) = 58

4.{10}

(a){2} Give a definition for Arrhenius acid and base

An acid increases concentration of H⁺ in aqueous solution

A base increases concentration of H⁺ in aqueous solution

(b) {2} Give a definition for Brønsted -Lowry acid and bas

An acid in aqueous solution donates proton

A base in aqueous solution accepts proton

(c) {2} Write down the equation for and value of K_w at 25^oC

K_w = [H₃O⁺][OH^-] = 1.0 x10^-14

(d) {4} at 25^oC, pH+ pOH= 14

pH= 7

pOH= 7

[H⁺] = 1.0 x10^-7

[OH^-]= 1.0 x10^-7

5. {5} Write down conjugate bases for the following acids:

HCl HCO₃^- OH^- CH₄ CH₃COOH

Cl^- CO₃^2- O^2- CH₃⁺ CH₃COO^-

6. {5} Calculate the pH of a solution at 25^OC, which contains 1.94 x 10^-10 M hydronium ions

-log1.94 x 10^-10 = 9.712

7. {5} Calculate the molarity of H₃O⁺ at 25⁰C in a solution with pOH=4.233

pH= 14-4.233= 9.767

[H⁺] = 10^-9.767 = 1.71 x 10^-10

8. {5} A 2.00-L aqueous solution contains 0.150 mol of HCl at 25⁰C. Calculate pH.

[H⁺] = [HCl] = 0.150 mol/2.00L = 0.0750 M

pH = -log 0.0750 = 1.125

9. {15} The pH of a 0.15M solution of the weak acid, HA, at 25⁰C is 5.35.

(a) {10} Calculate K_a of HA. [H⁺] = [A^-] = 10^-5.35 = 4.466 x10^-6

K_a =( = 4.466 x10^-6)²/(0.15- 4.466 x10^-6) = 1.33 x10^-10

(b) {5} Calculate the percent ionization of HA

% ionization = ( 4.466 x10^-6/0.15) x 100% = 0.003%

10. {10}. The K_a of hypochlorous acid, HClO, is 3.00x 10^-8. What is pH at 25.⁰C of a 0.0200M solution of HClO?

HA <- > H⁺ + A^-

K_a = [H⁺][A^-]/[HA]; x²/0.0200 = 3.00x 10^-8

X= (3.00x 10^-8 x 0.0200)^1/2 = (6.00 x 10^-10)^1/2= 2.449 x10^-5

pH = -log 2.449 x10^-5 = 4.611

11. {5} The pH of a 0.10M solution of a weak base is 10.0. What is K_b?

B(aq) + H₂O (l) ß> BH⁺ (aq) + OH^- (aq)

pOH =14-10=4; [OH] = 10 ^-4 ; K_b = (10^-4)²/0.1 = 1.0 x 10^-7