Chem 1310: Introduction to physical chemistry Part 3: Equilibria Solving concentration questions - ex 69.

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Presentation transcript:

Chem 1310: Introduction to physical chemistry Part 3: Equilibria Solving concentration questions - ex 69

CH 3 COOH (aq) ⇋ CH 3 COO - (aq) + H + (aq) ex 69 (b) CH 3 COOHCH 3 COO - H+H+ I111 C-x-x+x+x+x+x E1-x1+x Q C = 1 >> K C

CH 3 COOH (aq) ⇋ CH 3 COO - (aq) + H + (aq) ex 69 (b) CH 3 COOHCH 3 COO - H+H+ I111 I'200 C-x-x+x+x+x+x E2-xxx First let reaction go to reactants

H 2 O (aq) ⇋ H + (aq) + OH - (aq) ex 69 (a) H+H+ OH - I11 C+x+x+x+x E1+x Q C = 1 >> K C

H 2 O (aq) ⇋ H + (aq) + OH - (aq) ex 69 (a) First let reaction go to reactants H+H+ OH - I11 I'00 C+x+x+x+x Exx

N H 2 ⇋ 2 NH 3 ex 69 (c) Q C = 1 << K C N2N2 H2H2 NH 3 I111 C-x-x-3x+2x E1-x1-3x1+2x

N H 2 ⇋ 2 NH 3 ex 69 (c) First let reaction go to products N2N2 H2H2 NH 3 I111 I'2/305/3 C+0.5x+1.5x-x-x E2/3+0.5x1.5x5/3-x

2 O 3 ⇋ 3 O 2 ex 69 (d) First let reaction go to products O3O3 O2O2 I11 I'02.5 C+(2/3)x-x-x E(2/3)x2.5-x Q C = 1 << K C

2 NO 2 ⇋ N 2 O 4 ex 69 (e) First let reaction go to products NO 2 N2O4N2O4 I11 I'01.5 C+2x-x-x E2x2x1.5-x Q C = 1 < K C

HCOO - (aq) + H + (aq) ⇋ HCOOH (aq) ex 69 (f) First let reaction go to products Q C = 1 < K C HCOO - H+H+ HCOOH I111 I'002 C+x+x+x+x-x-x Exx2-x What % of the time is each HCOOH molecule ionized?

Ag + (aq) + I - (aq) ⇋ AgI (s) ex 69 (g) First let reaction go to products Q C = 1 << K C Ag + I-I- I11 I'00 C+x+x+x+x Exx