Chapter 17 equilibrium

Chapter 17 equilibrium - Chapter 17.1 17 Equilibrium: The...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
17 Chapter Equilibrium: The Extent of Chemical Reactions 17.1 Rate f = k f [R] Rate r = k r [P] K = f r k k = If the change is one of concentrations, the product concentration( s ) will increase, while the reactant concentration( s ) will decrease, but the ratio (equilibrium constant) will be constant. If the change is one of temperature, the product concentration( s ) and equilibrium constant will increase while the reactant concentration( s ) will decrease. 17.2 The faster the rate and greater the yield, the more useful the manufacturing process. 17.3 A system at equilibrium continues to be very dynamic at the molecular level. Reactant molecules continue to form products, but at the same rate that the products decompose to re-form the reactants. 17.4 A very large K eq means at equilibrium the value of product concentrations is much larger than the value of reactant concentrations, indicating that the reaction goes essentially to completion. 17.5 This reaction probably would have a large K , since it is very exothermic. 17.6 No. The value of Q is determined by the mass action expression with arbitrary concentrations for products and reactants. Thus, its value is not constant. 17.7 The equilibrium constant expression is K = [O 2 ]. If the temperature remains constant and the initial amount of Li 2 O 2 present was sufficient to reach equilibrium, the amount of O 2 obtained will be constant, regardless of how much Li 2 O 2 ( s ) is present. 17.8 a) H 2 ( g ) + I 2 ( g ) h 2HI( g ) Q = 2 [HI] [H ][I ] The value of Q increases as a function of time until it reaches the value of K . b) No 17.9 A homogeneous equilibrium reaction exists when all the components of the reaction are in the same phase (i.e., gas, liquid, solid, aqueous) 2NO( g ) + O 2 ( g ) h 2NO 2 ( g ) A heterogeneous equilibrium reaction exists when the components of the reaction are in different phases. Ca(HCO 3 ) 2 ( aq ) h CaCO 3 ( s ) + H 2 O( l ) + CO 2 ( g ) 54 [ ] [ ] P R
Background image of page 1

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

View Full DocumentRight Arrow Icon
17.10 1/2 N 2 ( g ) + 1/2 O 2 ( g ) h NO( g ) Q c(ref) = 1/2 1/2 2 2 [NO] [NO ] [O ] NO( g ) h 1/2 N 2 ( g ) + 1/2 O 2 ( g ) Q c = 1/2 1/2 2 2 [N ] [O ] [NO] Q c = 1/ Q c(ref) , so the constants do differ. 17.11 3/2 H 2 ( g ) + 1/2 H 2 ( g ) h NH 3 ( g ) Q c(ref) = 3 3/2 1/2 2 2 [NH ] [H ] [N ] 3 H 2 ( g ) + H 2 ( g ) h 2NH 3 ( g ) Q c = 2 3 3 [NH ] [H ] [N ] Q c = [ Q c(ref) ] 2 , so the constants do differ. 17.12 a) 4NO( g ) + O 2 ( g ) h 2N 2 O 3 ( g ) Q C = 2 2 3 4 [N O ] [NO] [O ] b) SF 6 ( g ) + 2 SO 3 ( g ) h 3 SO 2 F 2 ( g ) Q C = 3 2 2 2 [SO F ] [SF ][SO ] c) 2 SC1F 5 ( g ) + H 2 ( g ) h S 2 F 10 ( g ) + 2 HCl( g ) Q C = 2 2 10 2 [S F ][HCl] [SClF ] [H ] 17.13 a) 2 C 2 H 6 ( g ) + 7 O 2 ( g ) h 4 CO 2 ( g ) + 6 H 2 O( g ) Q c = 4 6 2 2 2 7 [CO ] [H O] [C H ] [O ] b) CH 4 ( g ) + 4F 2 ( g ) h CF 4 ( g ) + 4 HF( g ) Q c = 4 4 4 [CF ][HF] [CH ][F ] c) 2SO 3 ( g ) h 2SO 2 ( g ) + O 2 ( g ) Q c = 2 2 2 2 [SO ] [O ] [SO ] 17.14 a) 2NO 2 Cl( g ) h 2NO 2 ( g ) + Cl 2 ( g ) Q C = 2 2 2 2 [NO ] [Cl ] [NO Cl] 55
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 21

Chapter 17 equilibrium - Chapter 17.1 17 Equilibrium: The...

This preview shows document pages 1 - 3. Sign up to view the full document.

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