EquilibriumCh6-8

EquilibriumCh6-8 - Investigating Chemical Reactions N2O4(g...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
10/28/2011 1 Investigating Chemical Reactions •N 2 O 4 (g) 2 NO 2 (g) • Colorless brown Closed Container: Reversibility Groups of Molecules A state of Dynamic Equilibrium
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
10/28/2011 2 Equilibrium Concentrations • Equilibrium is reached when the concentrations of products and reactants remains unchanged with time • Condition dependent • Qualitative descriptions – “Equilibrium lies to the left (or right)” – “Equilibrium favors products (or reactants)” Initial: all NO 2 2 NO 2 N 2 O 4 Rates of Reaction • For simple, one step reaction, reaction rate is due to inherent reactivity and collision rate • Forward rate = k f [N 2 O 4 ] • Reverse rate = k r [NO 2 ] 2 Rates change over Course of Reaction Initial: all NO 2 2 NO 2 N 2 O 4
Background image of page 2
10/28/2011 3 Same Principle for all Reactions CO + H 2 O CO 2 + H 2 Law of Mass Balance General, Empirical Form: aA + bB cC + dD K eq = ሾ஼ሿ ሾ஽ሿ ሾ஺ሿ ሾ஻ሿ This is called the Equilibrium expression Law of Mass Balance • Empirical law—with justification from kinetics • Example: N 2 O 4 2 NO 2 = ሾேை ሾே = Equilibrium Constant Forward rate = reverse rate ݇ [N 2 O 4 ] = ݇ ሾܱܰ 2 This derivation is a simplification, but the outcome generally applies
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
10/28/2011 4 N 2 O 4 (g) 2 NO 2 (g) K eq = [NO 2 ] 2 /[N 2 O 4 ] If equilibrium [R] and [P] are known, K eq can be calculated. Example: at equilibrium, –[NO 2 ] = 1.50 x 10 -2 M, –[N 2 O 4 ] = 1.03 x 10 -2 M @ 317K (from experiment) K eq = (1.50 x 10 -2 ) 2 /(1.03 x 10 -2 ) = 0.0218 K eq depends on how the equation is written. •N 2 O 4 (g) 2 NO 2 (g) –K 1 = [NO 2 ] 2 /[N 2 O 4 ] = 0.0218 •2 NO 2 (g) N 2 O 4 (g) –K 2 = [N 2 O 4 ] / [NO 2 ] 2 –K 2 = 1/K 1 –K 2 = 1 / 0.0218 = 45.9 @ 317K •½ N 2 O 4 (g) NO 2 (g) –K 3 = [NO 2 ]/[N 2 O 4 ] 1/2 –K 3 = (K 1 ) 1/2 –K 3 = (0.0218) 1/2 = 0.148 Small Keq (less than 1) means less product at equilibrium Large Keq (more than 1) means more product at equilibrium Test Your Understanding Calculate the equilibrium constant for at a given temperature if a 1-L container that initially held 1 mol of CO and 1 mol of water reached equilibrium and then had 0.13 mol of CO 2 and H 2 . CO + H 2 O CO 2 + H 2 Answer: K eq
Background image of page 4
5 K eq describes ratio of reactants and products •A B Keq =0.33 • Which of the following systems are at equilibrium? – [A] = 3.0 M, [B] = 1.0 M – [A] = 7.5 x 10 -3 M, [B] = 2.5 x 10 -3 M – [A] = 12.0 M, [B] = 4.0 M •K eq describes ratio, not absolute concentrations Concentrations are dependent on starting point, Ratio of concentrations is not! Initial:
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 20

EquilibriumCh6-8 - Investigating Chemical Reactions N2O4(g...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online