Chemistry Review Manual 2
Chemistry Review Manual 2 Chemistry Review Manual 2
The equilibrium constant, K, for the reaction, c A A + c B B c Y Y + c Z Z is given by a a K = 5.2 a a CY Y CA A CZ Z CB B where a A , a B , a Y and a Z are the activities of A, B, Y and Z under equilibrium conditions, and c A , c B , c Y and c Z are the stoichiometric coefficients of A, B, Y and Z, respectively. Moreover, any combination of a A , a B , a Y and a Z satisfying this equation constitutes equilibrium conditions. The equilibrium constant depends only temperature. CX For more general reactions, there are additional a X factors in the numerator, if there are additional products; and in the denominator, if there are additional reactants. Examples are provided as follows: Reaction Equilibrium constant Br 2 (g) + Cl 2 (g) 2 BrCl(g) K = p p p 2 BrCl Br2 Cl2 H 2 (g) + I 2 (s) 2 HI(g) K = p p 2 HI H2 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K = p p 2 NH3 3 N p 2 H2 CaO(s) + H 2 O(l) Ca 2+ (aq) + 2 OH − (aq) K = [Ca ][OH ] 2+ − 2 HClO(aq) + H 2 O(l) ClO − (aq) + H 3 O + (aq) NaF(s) Na + (aq) + F − (aq) K = K a = K = K sp = + − + − [H3O ][ClO ] [H ][ClO ] [HClO] + − [Na ][F ] or [HClO] The last two examples show named equilibrium constants, K a and K sp , the acid ionization constant and the solubility product, respectively. In general, acids are characterized by the acid ionization constant, K a + − [H3O ][A ] = , 5.3 [ HA] the equilibrium constant for the acid ionization reaction, HA(aq) + H 2 O(l) H 3 O + (aq) + A − (aq) . Page 40 of 88
Bases are characterized by the base ionization constant, K b + − [HB ][OH ] = , 5.4 [ B] the equilibrium constant for the base ionization reaction, Salts are characterized by the solubility product, B(aq) + H 2 O(l) HB + (aq) + OH − (aq) . K sp c+ a a− c = [M ] [X ] 5.5 the equilibrium constant for the dissolution of the salt, M a X c (s) a M c+ (aq) + c X a− (aq) . Example 5.1: Write down the equilibrium constant, K, for each of the following reactions. (a) CaCO 3 (s) CaO(s) + CO 2 (g) (b) NH 4 HS(s) NH 3 (g) + H 2 S(g) (c) NO(g) + O 3 (g) NO 2 (g) + O 2 (g) (d) Cl 2 (aq) + H 2 O(l) H + (aq) + Cl − (aq) + HClO(aq) (e) Cl 2 (g) + 2 Fe 2+ (aq) 2 Cl − (aq) + 2 Fe 3+ (aq) Approach: Look for gases or aqueous species. The equilibrium constant is constructed from the partial pressures of gases and the concentrations of aqueous species, raised to the associated stoichiometric coefficients in the balanced chemical reaction. Products appear in the numerator, while reactants appear in the denominator. (a) K = p CO2 The equilibrium constant is just the partial pressure of carbon dioxide. K here is the vapour pressure of solid carbon dioxide. In an enclosed space, CO 2 (s) will sublimate until the partial pressure of CO 2 (g) equals K and the solid and gas are in equilibrium. (b) K = p p NH H S 3 2 NH 4 HS(s) does not appear in the equilibrium constant because it is a pure solid. All solids and liquids you will encounter in chemical reactions are pure (water, in aqueous reactions, is treated as though it were pure). (c) K = p NO2 O2 p NO p p O3 (d) + − [H ][Cl ][HClO] K = [Cl ] 2 Page 41 of 88
- Page 1 and 2: Chemistry Review Manual 2 nd Editio
- Page 3 and 4: Example 4.1: Type of bonding 31 4.2
- Page 5 and 6: 1. The Mole Early chemists discover
- Page 7 and 8: 2 Ca(s) + O 2 (g) → 2 CaO(s) It i
- Page 9 and 10: 1.5 Solutions Sometimes we are give
- Page 11 and 12: Problems: 1.1 How many grams of cal
- Page 13 and 14: 1.3 Mass of water produced = volume
- Page 15 and 16: The molar mass is determined from T
- Page 17 and 18: pV amount of H 2 consumed = n = RT
- Page 19 and 20: Solutions: 2.1 (a) These three gase
- Page 21 and 22: The amount of Al = (2/3) × 0.0043
- Page 23 and 24: 3.2 Strong and Weak Acids and Bases
- Page 25 and 26: Note that weak acids are conjugate
- Page 27 and 28: onds to the carbon atom. The result
- Page 29 and 30: Solutions: 3.1 (a) 2 Li(s) + Cl 2 (
- Page 31 and 32: 4. Bonding 4.1 Types of Bonding The
- Page 33 and 34: e.g. δ+ H H δ+ δ+ δ+ H H δ+ O
- Page 35 and 36: that pH = 7 tells us that the solut
- Page 37 and 38: Solutions: 4.1 (a) Cl 2 (g). The bo
- Page 39: 5. Chemical Equilibrium 5.1 Dynamic
- Page 43 and 44: Suppose we have a table of equilibr
- Page 45 and 46: Whether we add or remove product, t
- Page 47 and 48: from which we get 4 4 = = 4 2 ⎛ x
- Page 49 and 50: The solution of this quadratic equa
- Page 51 and 52: consider dissolving CaCO 3 (s) in a
- Page 53 and 54: which gives Using the quadratic for
- Page 55 and 56: (c) H 2 S K a1 = 8.9 × 10 −8 App
- Page 57 and 58: ( ) 2 − 1.8× 10 ± 1.8× 10 + 4
- Page 59 and 60: 5.2 An equilibrium mixture of N 2 O
- Page 61 and 62: Solutions: 5.1 (a) K = p p p 2 BrCl
- Page 63 and 64: 5.5 CH 4 (g) + Br 2 (g) CH 3 Br(g
- Page 65 and 66: (b) HOCl(aq) + H 2 O(l) H 3 O + (
- Page 67 and 68: 6. Thermochemistry Chemical reactio
- Page 69 and 70: or which rearranges to q water =
- Page 71 and 72: 6.4 Energy Conservation - the First
- Page 73 and 74: The number of moles of H + (aq) is
- Page 75 and 76: Note that an element in the standar
- Page 77 and 78: ∆Hgas phase reaction ≅ 2 D[C-C]
- Page 79 and 80: Solutions: 6.1 Heat is related to t
- Page 81 and 82: 3 3 × ⎯⎯→ × − 2 2 o −1
- Page 83 and 84: 7. Math Skills Chemistry is a quant
- Page 85 and 86: The value of the gas constant used
- Page 87 and 88: If log x < 0, then 0 < x < 1. Note
The equilibrium constant, K, for the reaction,<br />
c A A + c B B c Y Y + c Z Z<br />
is given by<br />
a a<br />
K = 5.2<br />
a a<br />
CY<br />
Y<br />
CA<br />
A<br />
CZ<br />
Z<br />
CB<br />
B<br />
where a A , a B , a Y and a Z are the activities of A, B, Y and Z under equilibrium conditions, and c A ,<br />
c B , c Y and c Z are the stoichiometric coefficients of A, B, Y and Z, respectively. Moreover, any<br />
combination of a A , a B , a Y and a Z satisfying this equation constitutes equilibrium conditions. The<br />
equilibrium constant depends only temperature.<br />
CX<br />
For more general reactions, there are additional a<br />
X<br />
factors in the numerator, if there are<br />
additional products; and in the denominator, if there are additional reactants. Examples are<br />
provided as follows:<br />
Reaction<br />
Equilibrium constant<br />
Br 2 (g) + Cl 2 (g) 2 BrCl(g)<br />
K<br />
=<br />
p<br />
p p<br />
2<br />
BrCl<br />
Br2 Cl2<br />
H 2 (g) + I 2 (s) 2 HI(g)<br />
K<br />
=<br />
p<br />
p<br />
2<br />
HI<br />
H2<br />
N 2 (g) + 3 H 2 (g) 2 NH 3 (g)<br />
K<br />
=<br />
p<br />
p<br />
2<br />
NH3<br />
3<br />
N<br />
p<br />
2 H2<br />
CaO(s) + H 2 O(l) Ca 2+ (aq) + 2 OH − (aq)<br />
K<br />
= [Ca ][OH ]<br />
2+ − 2<br />
HClO(aq) + H 2 O(l) ClO − (aq) + H 3 O + (aq)<br />
NaF(s) Na + (aq) + F − (aq)<br />
K = K a<br />
=<br />
K = K sp<br />
=<br />
+ −<br />
+ −<br />
[H3O ][ClO ] [H ][ClO ]<br />
[HClO]<br />
+ −<br />
[Na ][F ]<br />
or<br />
[HClO]<br />
The last two examples show named equilibrium constants, K a and K sp , the acid ionization<br />
constant and the solubility product, respectively.<br />
In general, acids are characterized by the acid ionization constant,<br />
K a<br />
+ −<br />
[H3O ][A ]<br />
= , 5.3<br />
[ HA]<br />
the equilibrium constant for the acid ionization reaction,<br />
HA(aq) + H 2 O(l) H 3 O + (aq) + A − (aq) .<br />
Page 40 of 88
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