Fundamentals of General Chemistry. Terms and Problems in Tests In 2 parts. P.1. Terms and Examples in Tasks. Study guide
.pdf7.The rate constant of a chemical reaction does not depend on: A) the nature of the reactants;
B) the presence of a catalyst;
C) the concentration of the reactants; D) the temperature.
8.The limiting stage of a multi-stage chemical reaction is:
A)the fastest stage;
B)the stage for which the activation energy is minimal;
C)the slowest stage;
D)the first stage.
9.What active particles are involved in chain reactions? A) cations and anions;
B) active molecules;
C) radicals and molecules; D) only free radicals.
10.The chemical reaction rate constant is equal to the reaction rate if: A) the reaction is heterogeneous;
B) the reaction is homogeneous;
C)the concentration of the reactants is 1 mol·L−1;
D)the concentrations of the reactants are the same.
11. The units of the rate constant for a second order reaction are:
A)mol · L–1 · s–1;
B)s–1;
C)mol · L · s–1;
D)L · mol–1 ·s–1.
8 . 4 . З а д а н и я д л я с а м о к о н т р о л я
1. В каталитической реакции катализатор: А. уменьшает энергию активации реакции; Б. изменяет константу равновесия реакции;
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В. изменяется в ходе химической реакции; Г. не изменяет механизма реакции.
2. Влияние катализатора на протекание реакции отражает кривая 2 рисунка:
А.
Б.
В. 
3. При повышении температуры от 60 до 80 °C скорость реакции увеличилась в 16 раз. Температурный коэффициент скорости реакции равен:
А) 8; Б) 2; В) 4; Г) 16.
4.Увеличение скорости реакции при введении катализатора связано: А) с уменьшением энергии активации нового пути реакции; Б) с увеличением энтальпии системы; В) с увеличением энергии активации нового пути реакции; Г) с увеличением давления в системе.
5.Если константа скорости первой реакции (k'), больше константы скорости второй реакции (k"), то соотношение между энергиями активации этих реакций:
А) Е'а > Е"а; |
Б) Е'а < Е"а; |
В) нельзя определить; |
Г) Е'а = Е"а. |
6. Правило Вант-Гоффа по температурной зависимости скорости реакции выражается уравнением:
А) k = A∙e–Еа⁄RT; |
Б) 2 = 1∙γ Т/10; |
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В) = k∙Ca∙Cb; |
Г) |
kТ + 50 |
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kТ |
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7. Константа скорости химической реакции не зависит: А) от природы исходных веществ; Б) от наличия катализатора;
В) от концентрации исходных веществ; Г) от температуры.
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8.Лимитирующей стадией многостадийной химической реакции является:
А) самая быстрая стадия; Б) стадия, для которой энергия активации минимальна; В) самая медленная стадия; Г) первая стадия.
9.При участии каких активных частиц происходят цепные реакции? А) катионов и анионов; Б) активных молекул; В) радикалов и молекул;
Г) только свободных радикалов.
10.Константа скорости химической реакции равна скорости реакции, если:
А) реакция гетерогенная; Б) реакция гомогенная;
В) концентрации исходных веществ равны 1 моль/л; Г) концентрации исходных веществ одинаковы.
11.Размерность константы скорости для реакции второго порядка:
А) моль · л–1 · с–1; Б) с–1; В) моль · л · с–1;
Г) л · моль–1 ·с–1.
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9 . H Y D R O L Y S I S
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9 . 1 . D i c t i o n a r y |
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English |
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Anion hydrolysis |
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Гидролиз по аниону |
Aqueous |
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Водный |
Blue |
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Синий |
Cation hydrolysis |
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Гидролиз по катиону |
Сolourless |
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Бесцветный |
Degree of hydrolysis |
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Степень гидролиза |
Dilute |
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Разбавленный |
Dilution |
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Разбавление |
Distilled |
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Дистиллированная |
First step |
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Первая cтупень |
Indicator |
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Индикатор |
Insoluble |
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Нерастворимый |
Irreversible |
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Необратимый |
Hydrolysis |
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Гидролиз |
Hydrolysis constant |
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Константа гидролиза |
Litmus |
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Лакмус |
Methyl orange |
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Метиловый оранжевый |
Orange |
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Оранжевый |
Pink-purple |
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Малиновый |
Phenolphthalein |
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Фенофталеин |
Red |
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Красный |
Reversible |
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Обратимый |
Shift |
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Сдвиг |
Soluble |
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Растворимый |
Yellow |
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Желтый |
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9 . 2 . W o r k e d e x a m p l e s
Example 1. Write the equation of copper (II) nitrate hydrolysis in ionic and molecular forms. What color is the litmus in this solution? Give the expression of hydrolysis constant of the first step for the ions Сu(II).
Answer
The copper nitrate is a salt formed by the weak base (Cu(OH)2) and the strong acid (HNO3). The Cu(NO3)2 salt undergoes the cation hydrolysis. The first hydrolysis step equations are
Cu2+ + HOH CuOH+ + H+ (the ionic molecular form);
Cu(NO3)2 + H2O CuOHNO3 + HNO3 (the molecular form).
An excess of hydrogen ions yields an acidic solution (pH < 7). The litmus in acidic solution is red (table 9.1).
The chemical equilibrium constant expression for the first hydrolysis step is:
[CuOH+][H+]
K = [Cu2+][H2O]
or
K[H2O] = [CuOH+][H+].
The product K[H2O] is a new constant and is called the hydrolysis constant Kh1. Then constant expression of the first hydrolysis step for the ions Cu2+ is:
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[CuOH+][H+] |
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Kh1 |
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[Cu2+ |
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Example 2. For Н2S, Kа1 = 1.0 10−7, Kа2 = 2.5 10−13. Taking into account that the anion hydrolysis is limited by the first step, calculate the pH of a 0.1 M Na2S solution.
Answer
The Na2S is formed by the strong base (NaOH) and the weak dibasic acid (Н2S). This salt undergoes the anion hydrolysis. The first hydrolysis step proceeds as
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S2− + HOH HS− + OH−. |
(9.1) |
An excess of hydroxide ions yields a basic solution.
The salt hydrolysis constant (by anion or cation) is calculated using the following equations:
I hydrolysis step
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KH |
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KH |
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Kh1 (by cation) = |
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Kh1 (by anion) = |
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Kb3 |
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Ka3 |
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II hydrolysis step |
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KH |
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KH |
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Kh2 (by cation) = |
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Kh2 (by anion) = |
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Kb2 |
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Ka2 |
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III hydrolysis step |
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KH |
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KH |
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Kh3 (by cation) = |
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Kh3 (by anion) = |
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Kb1 |
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Ka1 |
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where Kb1, Kb2, Kb3 и Ka1, Ka2, Ka3 are step dissociation constant of base and acid, respectively.
The value of hydrolysis constant in the first step for the ions S2− is:
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[HS−][OH−] |
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KH |
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10−14 |
10−2. |
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Kh1 |
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[S |
2− |
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Kа2 |
2,5 10 |
−13 |
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From equation (9.1), [HS−] = [OH−]. We assume that [S2−] is equal to the initial Na2S concentration in solution (Csalt = 0.1 M), so
[OH−]2
Kh1 = [S2−] ,
then
[OH−] = √Kh1Csalt = √4.0 10−2 0.1 = 6.32 10−2.
To determine the pH, use the formulas:
рН = 14 − рОН; рОН = −lg[OH−].
Then
рОН = −lg(6.32 10−2) = 1.2 and рН = 14 − 1.2 = 13.8.
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Example 3. For Ni(OН)2, Kb2 = 8.32 10−4. Taking into account that the anion hydrolysis is limited by the first stage, calculate the pH of a 0.1 M NiSO4 solution.
Answer
The NiSO4 salt is formed by the weak base (Ni(OН)2) and the strong acid (Н2SO4). This salt undergoes the cation hydrolysis. The first hydrolysis step proceeds as:
Ni2+ + HOH NiOH+ + H+. |
(9.2) |
An excess of hydrogen ions yields an acidic solution. The value of hydrolysis constant in the first step for Ni2+ is:
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[NiOH+][H+] |
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KH |
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10−14 |
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10−11. |
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Kh1 |
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Kb2 |
8.32 10 |
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From equation (9.2), [NiOH+] = [H+]. Assume that [Ni2+] is equal to the initial concentration of NiSO4 in the solution (Csalt = 0.1 M), so
[H+]2
Kh1 = [Ni2+] ,
then
[H+] = √Kh1Csalt = √1.2 10−11 0.1 = 1.1 10−6.
To determine the pH, use the formula
рН = −lg[H+] = −lg(1.1 10−6) = 5.96.
Table 9.1
The indicator colour at different pH of solution
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pH range of |
The indicator colour in solution |
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Indicator |
indicator |
more acidic, |
more basic, |
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colour |
рН < |
рН > |
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change |
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Methyl orange |
3.1 – 4.4 |
< 3.1 |
> 4.4 |
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Red |
yellow |
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End of table 9.1 |
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pH range of |
The indicator colour in solution |
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Indicator |
indicator |
more acidic, |
more basic, |
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colour |
рН < |
рН > |
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change |
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Methyl red |
4.2 – 6.2 |
< 4.2 |
> 6.2 |
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Red |
yellow |
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Litmus |
5.0 – 8.0 |
< 5,0 |
> 8.0 |
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Red |
blue |
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Phenolphtha- |
8.0–10.0 |
< 8.0 |
> 10.0 |
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colourless |
pink-purple |
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9 . 3 . T a s k s f o r s e l f - c o n t r o l
1. The degree of aluminium сhloride hydrolysis decreases by adding to
the solution of: |
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A) sodium hydroxide; |
B) sulfuric acid; |
C) sodium сhloride; |
D) distilled water. |
2. In aqueous solution of Na2S the hydrolysis of sulfide ions S2− + HOH HS− + OH−
increases at:
A)dilution of the solution;
B)heating of the solution;
C)adding to the solution of aqueous alkali;
D)adding to the solution of an acid.
3.The degree of the lead nitrate hydrolysis increases at: A) dilution of the solution;
B) heating of the solution;
C) adding to the solution of dilute nitric acid; D) adding to the solution of aqueous alkali.
4.Find a relationship between the nature of the salt and products of its hydrolysis in the first step:
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1. AlCl3 |
А) [AlOH]2+ + H+ |
2. K3PO4 |
В) Al(OH)3 + H+ |
3. K2 S |
C) HS− + OH− |
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D) H2S + OH− |
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F) HPO42− + OH− |
G)H2PO4− + OH−
5.Phenolphthalein colour in an aqueous solution of sodium carbonate is:
А) pinkpurple; |
B) blue; |
С) colourless; |
D) red. |
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6. |
Litmus turns red in an aqueous solution of the salt: |
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А) NaNO3; |
B) KI; |
C) ZnSO4; |
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D) K3PO4. |
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The solution is acidic, when the salt dissolved in water is: |
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А) CuCl2; |
B) NaCl; |
C) Na2S; |
D) Ca(NO3)2. |
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8. Find a relationship between the indicator and its colour in a neutral solution:
1. litmus
2. methyl orange
3. phenolphthalein
G)colourless.
9.The solution is basic, when the salt dissolved in water is:
А) Na2SO3; |
B) CuCl2; |
C) Na2SO4; |
D) Ba(NO3)2. |
10. For (Fe(OH)3, Kb2 = 1.82 10−11, Kb3 = 1.35 10−12. The value of the first step hydrolysis constant for Fe3+ is:
А) 5.5 10−4;
B)7.4 10−1;
C)7.4 10−3;
D)1.35 10−2.
11. For (Fe(OH)3, Kb2 = 1.2 10−11, Kb3 = 1.35 10−12. The value of the second step hydrolysis constant for Fe3+ is:
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А) 5.5 10−2;
B)7.4 10−3;
C)1.82 10−3;
D)5.5 10−4.
12. For Н3PO4, Kа1 = 7.1 10−3, Kа2 = 6.2 10−8, Kа3 = 5.0 10−13. The value of the second step hydrolysis constant for PO34– is:
A)1.4 10−12;
B)6.2 10−4;
C)2.0 10−2;
D)1.6 10−7.
9 . 4 . З а д а н и я д л я с а м о к о н т р о л я
1. Степень гидролиза хлорида алюминия уменьшается при добавлении
в раствор: |
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А) гидроксида натрия; |
Б) серной кислоты; |
В) хлорида натрия; |
Г) дистиллированной воды. |
2. В водном растворе сульфида натрия гидролиз сульфид-ионов
S2− + HOH HS− + OH−
возрастает: |
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А) при разбавлении раствора; |
Б) при повышении температуры; |
В) при добавлении в раствор щелочи; |
Г) при подкислении раствора. |
3.Степень гидролиза нитрата свинца возрастает: А) при разбавлении раствора; Б) при нагревании раствора;
В) при добавлении в раствор разбавленной азотной кислоты; Г) при добавлении в раствор щелочи.
4.Установите соответствие между природой соли и продуктами ее гидролиза по первой ступени:
1. AlCl3 |
А) [AlOH]2+ + H+; |
2. K3PO4 |
Б) Al(OH)3 + H+; |
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