Решение задач по химии

Finding the relative molecular mass of the substance .

1. Find the molar mass of aluminum sulfate . Indicate the relative molecular weight of the substance.

given:

АІ₂(SО₄)₃

find:

М[АІ₂(SО₄)₃] -?

Мr[АІ₂(SО₄)₃] -?

Solution:

М[АІ₂(SО₄)₃] = А(АІ) · n(АІ) + А(S) · n(S) + А(О) · n(О)

М[АІ₂(SО₄)₃] = 27 · 2 + (32 + 16 · 4) · 2 = 54 + 96 · 3 =

= 54 + 288 = 342 g/mol

Мr[АІ₂(SО₄)₃] = Аr(АІ) · n(АІ) + Аr(S) · n(S) + Аr(О) · n(О)

Мr[АІ₂(SО₄)₃] = 27 · 2 + (32 + 16 · 4) · 2 = 54 + 96 · 3 =

= 54 + 288 = 342 .

Answer: М[АІ₂(SО₄)₃] = 342 g/mol; Мr[АІ₂(SО₄)₃] = 342.

2. Find the molar mass of phosphoric acid. Indicate the relative molecular weight of the substance.

given:

Н₃РО₄

find:

М(Н₃РО₄) -?

Мr(Н₃РО₄) -?

Solution:

М(Н₃РО₄) = А(Н) · n(Н) + А(Р) · n(Р) + А(О) · n(О)

М(Н₃РО₄) = 1 · 2 + 31· 1 + 16 · 4 = 2 + 31 + 64 = 98 g/mol

Мr(Н₃РО₄) = Аr(Н) · n(Н) + Аr(Р) · n(Р) + Аr(О) · n(О)

Мr(Н₃РО₄) = 1 · 2 + 31· 1 + 16 · 4 = 2 + 31 + 64 = 98

Answer: М(Н₃РО₄) = 98 g/mol; Мr(Н₃РО₄) = 98.

Calculations by chemical formulas.

1. What is the mass attitude of elements in substances the formula which SO₃?

given:

SO₃

find:

m (S): m (O)

Solution:

Mr(SO₃) = Ar(S)+ 3Ar(O) = 32 + 3 · 16 = 80

Answer: m(S) : m(О) = 2 : 3.

2. What is the mass attitude of elements in substances the formula which CuO?

given:

CuO

find:

m(Cu) : m(О)

Solution:

Мr(CuO) = Аr(Cu) + Аr(O) = 64 + 16 = 80

Answer :m(Cu) : m(О) = 4 : 1.

Of the mass fraction of the chemical element in the compound .

1. Calculate the mass fraction of each element in marble , whose composition corresponds to the formula CaCO₃.

given:

CaCO₃

find:

ω(Ca) - ?

ω(C) -?

ω(O) -?

Solution:

Мr(CaCO₃) = Аr(Ca) + Аr(C) + 3Аr(O) = 40 + 12 + 3 · 16 = 100

Answer :ω(Ca) = 40 %; ω(C) = 12 %; ω(O) = 48 %.

2. What is the mass fraction of oxygen in the compound whose formula Al₂O₃?

given:

Al₂O₃

find:

ω(O) -?

Solution:

Мr(Al₂O₃) = 2Аr(Al) + 3Аr(O) = 2 · 27 + 3 · 16 = 102

Answer : ω(O) = 47 %.

3. Calculate the mass fraction of each element in the magnesium sulphate MgSO₄

given:

MgSO₄

find:

ω(Mg) - ?

ω(S) -?

ω(O) -?

Solution:

Мr(MgSO₄) = Аr(Mg) + Аr(S) + 4Аr(O) = 24 + 32 + 4 · 16 = 120

Answer:ω(Mg) = 20 %; ω(S) = 26,7 %; ω(O) = 53,3 %.

Derivation of the formulas of compounds .

1. Mass fractions of iron and sulfur of compounds are, respectively, 46.67% and 53.33 %.Determine the formula of this compound.

given:

ω(Fе) = 46,67 %

ω(S) = 53,33 %

find :

Fе_х S_у -?

Solution:

М(Fе) = 56 g/mol,

М(S) = 32 g/mol,

ν(Fе) : ν(S) = х : у = 0,83 : 1,66

formula iron compound FеS₂.

Answer: FеS₂.

2. Mass fraction of nitrogen in the of oxide nitrogen equal to 36.84 %. Output the simplest formula this of oxide .

given:

ω(N) = 36,84 %

find :

N_хО_у -?

Solution:

m(О) = m(N_хО_у) - m(N) = 100 g -36,84 g = 61,16 g

М(N) = 14 g/mol,

М(О) = 16 g/mol,

ν(N) : ν(О) = х : у = 2,63 : 3,95

simplest formula oxide N₂О₃.

Answer: N₂О₃.

Major quantitative characteristics substances: a substance amount , mass, volume and number of molecules.

1. Calculate the amount of a substance that is 120 g of sodium hydroxide.

given:

m(NаОН) = 120 g

find:

ν(NаОН) - ?

Solution:

М(NаОН) = 23 + 16 + 1 = 40 g/ mol

Answer:ν(NаОН) =3 mol.

2. Calculate the amount of substance which corresponds to 49 g of copper hydroxide (II).

given:

m(Сu(ОН)₂) = 49 g

find:

ν(Сu(ОН)₂) - ?

Solution:

М(Сu(ОН)₂) = 64 + 16 · 2 + 1 · 2 = 98 g/mol

Answer:ν(Сu(ОН)₂) =0,5 mol.

3. Calculate the amount of substance corresponding to 490 g of sulfuric acid, H₂SO₄ .

given:

m(Н₂SО₄) = 490 г

Найти:

ν(Н₂SО₄) - ?

Решение:

М(Н₂SО₄) = 1· 2 + 32 + 16 · 4 = 98 г/моль

Ответ: ν(Н₂SО₄) =.5 моль.

4. Calculate the amount of substance corresponding to 9.8 g phosphoric acid H₃PO₄ .

given:

m(Н₃РО₄) = 9,8 г

Найти:

ν(Н₃РО₄) - ?

Решение:

М(Н₃РО₄) = 1· 3 + 31 + 16 · 4 = 98 г/моль

Ответ: ν(Н₃РО₄) =.0,1 моль.

5. Equal amounts of whether the substance contains 4 g of hydrogen gas and oxygen gas of 64 g ?

given:

m(Н₂) = 4 g

m(О₂) = 64 g

find:

ν(Н₂) - ?

ν(О₂) - ?

Solution:

1- way:

М(Н₂) = 2 g/mol

4 g Н₂ ----- х mol Н₂

2 g Н₂ ----- 1 mol Н₂

2- way:

М(Н₂) = 2 g/mol

1- way:

М(О₂) = 32 g/mol

64 g О₂ ----- х mol О₂

32 g О₂ ----- 1 mol О₂

2- way:

М(О₂) = 32 g/mol

Answer: ν(Н₂) = 2 mol; ν(О₂) = 2 mol - Identical.

6. Identical to the number of substances contained in water, 18 and 17 g H₂S?

given:

m(Н₂О) = 18 g

m(H₂S) = 17 g

find:

ν(Н₂О) - ?

ν(H₂S) - ?

Solution:

1- way:

М(Н₂О) = 18 g/mol

18 g Н₂О ----- х mol Н₂О

18 g Н₂О ----- 1 mol Н₂О

2- way:

М(Н₂О) = 18 g/mol

1- way:

М(H₂S) = 34 g/mol

17g H₂S ----- х mol H₂S

34 g H₂S ----- 1 mol H₂S

2- way:

М(H₂S) = 34 g/mol

Answer:ν(Н₂О) = 1 mol; ν(H₂S) = 0,5 mol - Misc.

7. Identical do the amount of the substance contained in 44 g of CO₂ and

2 g of hydrogen H₂?

given:

m(СО₂) = 44 g

m(Н₂) = 4 g

find:

ν( СО₂) - ?

ν(Н₂) - ?

Solution:

1- way:

М(СО₂) = 44 g/mol

44 g СО₂ ----- х mol СО₂

44 g СО₂ ----- 1 mol СО₂

2- way:

М(СО₂) = 44 g/mol

1- way:

М(Н₂) =2 g/моль

2 g Н₂ ----- х mol Н₂

2 g Н₂ ----- 1 mol Н₂

2- way:

М(Н₂) = 2 g/mol

Answer:ν(СО₂) = 1 mol; ν(Н₂) = 1 mol-Identical.

8. What proportion of moles are 9 grams of water ?

given:

m(Н₂О) = 9 g

find:

ν(Н₂О) - ?

Solution:

1- way:

М(Н₂О) = 18 g/mol

9 g Н₂О ----- х mol Н₂О

18 g Н₂О ----- 1 mol Н₂О

2- way:

М(Н₂О) = 18 g/mol

Ответ: ν(Н₂О) = 0,5 mol.

9. What proportion of moles is 1 g of hydrogen H₂?

given:

m(Н₂) = 1 g

find:

ν(Н₂) - ?

Solution:

1- way:

М(Н₂) =2 g/mol

1 g Н₂ ----- х mol Н₂

2 g Н₂ ----- 1 mol Н₂

2- way:

М(Н₂) = 2 g/mol

Answer: ν(Н₂) = 0,5 mol.

10. What proportion of mole are 71 g of sodium sulfate Na₂SO₄?

given:

m(Na₂SO₄) = 71 g

find:

ν(Na₂SO₄) - ?

Solution:

1- way:

М(Na₂SO₄) = 142 g/mol

71 g Na₂SO₄ ----- х mol Na₂SO₄

142 g Na₂SO₄ ----- 1 mol Na₂SO₄

2- way:

М(Na₂SO₄) = 2 g/mol

Answer:ν(Na₂SO₄) = 0,5 mol.

The relative density of the gases.

1. Determine the relative density of nitrogen to hydrogen .

given:

М(N₂) = 28 g/mol

find:

Dн₂(N₂) -?

Solution:

М(Н₂) = 2 g/mol

Answer:Dн₂(N₂) = 14.

2. Determine the relative density of helium to hydrogen .

given:

М(Не) = 4 g/mol

find:

Dн₂(Не) -?

Solution:

М(Н₂) = 2 g/mol

Answer:Dн₂( Не) = 2.

3. Determine the relative density of butane C₄H₉ hydrogen

given:

М(С₄Н₉) = 57 g/mol

find:

Dн₂(С₄Н₉) -?

Solution:

М(Н₂) = 2 g/mol

Answer:Dн₂(С₄Н₉) = 29.

Mass , volume fraction of the substance in the mixture.

1. Determine the weight fraction of nitrogen in the mixture thereof with ammonia , if the volume fraction of nitrogen in the mixture is 10%.

given:

φ(N₂) = 10 %

find:

ω(N₂) -?

Solution:

V_mixture = 100 lit

V(N₂) = 10 lit, а V(NН₃) = 100 lit - 10 lit= 90 lit

М(N₂) = 28 g/mol

М(NН₃) = 17 g/mol

V_m= 22,4 l/mol

m_mixture = 12,5 g + 68,3g = 80,8 g

Answer:ω(N₂) = 0,155.

2. The powder mixture of magnesium and silver weight 1,5 g treated with excess of a solution of sulfuric acid ,wherein stood 560 ml (normal conditions) of hydrogen . Calculate the mass ratio ( in percentage ) of the components in the initial mixturegiven:

given:

m(mixture) =1,5 g

V (Н₂) = 560 ml

find:

ω(Мg) -?

ω(Аg) -?

Solution:

Мg + Н₂SО₄ = МgSО₄ + Н₂↑

Silver does not react :

Аg + Н₂SО_4(dil.) ↛

ν(Н₂) = ν(Мg) ⇒ν(Мg) = 0,025 mol

М(Мg) = 24 g/mol

m(Мg) = ν(Мg) · М(Мg) = 0,025 mol · 24 g/mol = 0,6 g

m(Аg) = m(mixture) - m(Мg) = 1,5 g - 0,6 g = 0,9 g

Answer:ω(Мg) = 40 %; ω(Аg) = 40 %.

3. A mixture of zinc oxide and zinc weight 8,525 g was treated with excess hydrochloric acid, wherein stood 2,24 liters of gas (normal conditions ) .Calculate the weight fraction (in percent ) of zinc oxide in the initial mixture.

given:

m(mixture) = 8,525 g

V(Н₂) = 2,24 lit

find:

ω(ZnО) -?

Solution:

ZnО + 2НСl = ZnСl₂ + 2Н₂О (1)

Zn + 2НСl = ZnСl₂ + Н₂↑ (2)

ν(Н₂) = ν(Zn) ⇒ ν(Zn) 0,1 mol

М(Zn) = 65 g/mol

m(Zn) = ν(Zn) · М(Zn) = 0,1 mol · 65 g/mol = 6,5 g

m(ZnО) = m(mixture) - m(Zn) = 8,525 g - 6,5 g = 2,065 g

Answer:ω(ZnО) = 24 %.

4. Determine the volume fraction of nitrogen in the mixture thereof with ammonia , if the mass fraction of nitrogen in the mixture is 62,2 %.

given:

ω(N₂) = 62,2 %

find:

φ(N₂) -?

Solution:

m_mixture = 100 g

m(N₂) = 62,2 g, а m(NН₃) = 100 g - 62,2 g = 37,8 g

М(N₂) = 28 g/mol

М(NН₃) = 17 g/mol

V_m= 22,4 l/mol

V_mixture = 49,76 lit + 50,47 lit = 100,23 lit

Answer:φ(N₂) = 0,50.

5. Calculatevolume fractionhydrogen and helium in a mixturecontaining 20 %( by weight)hydrogen .

given:

ω(Н₂) = 20 %

find:

φ(Н₂ and Не) -?

Solution:

М(Н₂) = 2 g/mol

m(Н₂) = ν(Н₂) · М(Н₂) = 20 mol · 2 g/mol = 40 g

М(N₂) = 28 g/mol

М(NН₃) = 17 g/mol

V_m= 22,4 л/mol

V = 49,76 + 50,47 = 100,23 lit

Answer:φ(N₂) = 0,50.

Electrolytic dissociation . Reactions in electrolyte solutions.

1. In the solution was placed of 100 molecules . Determine the degree of dissociation , if dissociated 5of molecules.

given:

n = 5 molecules

N = 100 molecules

find:

α -?

Solution:

Answer:α = 0,05.

2. In the solution was placed 100 molecules .Determine the degree of dissociation , if dissociated 15 of molecules .

given:

n = 15 molecules

N = 100 molecules

find:

α -?

Solution:

Answer:α = 0,15.

3. In the solution was placed 100 molecules .Determine the degree of dissociation , if dissociated 50 of molecules .

given:

n = 50 molecules

N = 100 molecules

find:

α -?

Solution:

Answer:α = 0,5.

4. In the solution was placed 100 molecules .Determine the degree of dissociation , if dissociated 95 of molecules .

given:

n = 95 molecules

N = 100 molecules

find:

α -?

Solution:

Answer:α = 0,95.

5. In 1 liter of water dissolved 56 l (under normal conditions) ofhydrogen chloride . In The resulting solution contains 13,244 · 10²³ of chloride ions. Determine the molar concentration of and the degree of dissociation of the obtained hydrochloric acid, if the density of the resulting solution is 1 g/ml.

given:

V(Н₂О) = 1 lit

V(НСl) = 56 lit

N(Сl^-) = 13,244· 10²³

ρ_solution(НСl) = 1,04 g/ml

find:

α(НСl) -?

с(НСl) -?

Solution:

N(НСl) = ν(НСl) · N_А = 2,5 mol · 6,02 · 10²³ = 15,05 · 10²³

НСl → Н⁺ + Сl^-

N(Сl^-) = N_dis(НСl) = 13,244· 10²³

М(НСl) = 36,5 g/mol

m(НСl) = ν(НСl) · М(НСl) = 2,5 mol · 36,5 g/mol = 91,2 V(Н₂О) · ρ(Н₂О) = 1000 ml · 1 g/ml = 5 g.

m(Н₂О) =1000 g.

m_solution(НСl) = m(Н₂О) + m(НСl) = 1000 g + 91,25 g = 1091,25 g.

Answer: ω(НСl) = 88 %; с(НСl) = 2,38 mol/lit.

6. The degree of dissociation of barium hydroxide on the first stage-92%, the second stage-56%. Calculate the number of cations barium and hydroxide ions in 0,5 liters 1,5 M solution.

given:

α₁(Ва(ОН)₂) = 92 %

α₂(Ва(ОН)₂) = 56 %

V_solution(Ва(ОН)₂) = 0,5 л = 500 ml

с(Ва(ОН)₂) = 1,5 М = 1,5 mol/lit

find:

N(Ва²⁺) -?

N(ОН^-) -?

Solution:

1. Ва(ОН)₂ ⇄ Ва(ОН)⁺ + ОН^-;

2. Ва(ОН)⁺⇄ Ва²⁺ + ОН^-.

ν(Ва(ОН)₂) = с(Ва(ОН)₂) · V_solution(Ва(ОН)₂) = 1,5 mol/lit · 0,5 lit = 0,75 mol

ν(Ва(ОН)⁺) = ν₁(ОН)^- = ν_dis(Ва(ОН)₂) = 0,69 mol

ν(Ва²⁺) = ν₂(ОН)^- = ν_dis(Ва(ОН)^⁺) = 0,386 mol

N(Ва²⁺) = ν(Ва²⁺) · N_А = 0,386 mol · 6,02 ·10²³mol^-1 = 2,324 ·10²³

ν(ОН^-) = ν₁(ОН)^- + ν₂(ОН)^- = 0,69 mol+ 0,386 mol = 1,076 mol

N(ОН^-) = ν(ОН^-) · N_А = 1,076 mol · 6,02 ·10²³mol^-1 = 6,478 ·10²³

Answer: N(Ва²⁺) = 2,324 ·10²³; N(ОН^-) = 6,478 ·10²³.

The chemical reaction rate . Chemical equilibrium .

1. When the temperature rises at a rate of 10⁰C chemical reaction increases two fold. When the 20⁰C it is equal to 0,04 mol/(lit · h). What will be the rate of the reaction at 40 ° C?

given:

γ = 2

υ₁ = 0,04 mol/lit· h

t₁ = 20⁰С

t₂ = 40⁰С

find:

υ₂ -?

Solution:

Answer: υ₂ = 0,16 mol/lit · h

2. When the temperature rises at a rate of 10⁰C chemical reaction increases two fold. When the 20⁰C it is equal to 0,04 mol/(lit · h). What will be the rate of the reaction at 10 ° C?

given:

γ = 2

υ₁ = 0,04 mol/lit · h

t₁ = 20⁰С

t₂ = 10⁰С

find:

υ₂ -?

Solution:

Answer: υ₂ = 0,02 mol/lit · h

3. At temperature 20 ° C the reaction lasts 2 hours 40 minutes. At what the reaction stops end for 5 minutes if the temperature coefficient of the reaction of is 2 .

given:

γ = 2

τ₁ = 2 h 40 min

τ₂ = 5 min

t₁ = 20⁰С

find:

t₂ -?

Solution:

Δt = t₂ - t₁

t₂ = Δt + t₁

1) τ₁ = 2 h 40 min = 2 · 60 + 40 = 160 min

τ₂ = 5 min

160 : 5 = 32 .

4) t₂ = 50 + 20 = 70

Answer: at 70⁰ ° C, the reaction is over in 5 minutes.

4. When the temperature rises at a rate of 100C chemical reaction increases two fold. When 20⁰C it is equal to

0,04 mol/( lit · h). What will be the rate of the reaction at 0 ° C?

given:

γ = 2

υ₁ = 0,04 mol/lit· h

t₁ = 20⁰С

t₂ = 0⁰С

find:

υ₂ -?

Solution:

Answer: υ₂ = 0,01 mol/lit · h

5. The reaction proceeds at a temperature of 50⁰C for

3 min 20 s.The temperature coefficient of the reaction rate is equal to 3 .How much time will end this reaction at 30⁰C .

given:

γ = 3

τ₁ = 3 min 20 s

t₁ = 50⁰С

t₂ = 30⁰С

find:

τ₂ -?

Solution:

τ inversely proportionalt⁰ С

In 9 times the rate of reaction will decrease as the temperature decreases from 50⁰C to 30⁰C

2) Time of course of the reaction at lower temperatures will increase by

9 times :200 sec · 9 = 1800 sec = 30 минут - за столько времени закончится эта реакция при понижении температуры с 50⁰С до 30⁰С.

Answer: at a temperature of 300 ° C for 30min the reaction stops.

6. Calculate of the average rate of the reaction : А + В = С, if after 10 seconds, the concentration of substance A was 0,2 mol/lit, and of after 40 seconds , the concentration of this substance was of 0,02 mol/lit.

given:

t₁ = 10 sec

t₂ = 40 sec

с₁ = 0,2 mol/lit

с₂ = 0,02 mol/lit

find:

υ - ?

Solution:

Answer: υ =0,06 mol/lit · sec.

7. 20 seconds after the start of the reaction 2SО₂ + О₂ → 2SО₃ 3,2 g remain unoxidised of 4,8 g of sulfur oxide (IV). Find the rate of oxidation of sulfur dioxide .

given:

m₁(SО₂) = 4,8 g

m₂(SО₂) = 3,2 g

Δt = 20 sec

find:

υ - ?

Solution:

М(SО₂) = 64 g/mol

if V = 1 lit, т.

Answer: υ =0,00125 mol/lit · sec.

8. The decomposition reaction HBr on simple substance proceeds in the vessel volume 2 liters.Initially, the in vessel containing the 0,5 mol/lit of hydrogen bromide , the after 20 seconds the amount substance was 0,3 mol / l,the after 40 seconds 0,1 mol/lit, and after 1 minute 0,05 mol/lit. Calculate the average of speed response of to the three time phases.

given:

Δt₁ = 20 sec

Δt₂ = 40 sec

Δt₂ = 1 min = 60 sec

с₁(НВr) = 0,5 mol/lit

с₂(НВr) = 0,3 mol/lit

с₃(НВr) = 0,1 mol/lit

с₄(НВr) = 0,05 mol/lit

find:

υ₁ - ?

υ₂ - ?

υ₃ - ?

Solution:

Answer: υ₁ = 0,01 mol/lit · s

υ₂ = 0,05 mol/lit · s

υ₃ = 0,0008 mol/lit· s

9. After a certain period of time after the start of the reaction equation which СО_2(g) + С_(s) = 2СО_(s), concentration of carbon oxide is reduced by 4 times. how many times this will reduce the reaction rate compared to the initial ?

given:

find:

Solution:

Answer: the reaction rate decreases 16 times .

10. A chemical reaction occurs in the solution according to the following equation: А + В = С. Initial concentrations :

A - 2 mol/lit, B - 2,5 mol/lit. After 20 minutes the concentration of A decreased to 1 mol/lit. What was of concentration of the substance in ?Determine the average rate of the reaction .

given:

Δt = 20 min

С₁(А) = 2 mol/lit

С₂(А) = 1 mol/lit

find:

С₃(В) -?

υ₁ - ?

Solution:

20 min = 20· 60= 1200 s

А + В = С

С₁(А) = с₁(В) = 2 mol/lit

С₂(А) = с₂(В) = 1 mol/lit

С₂(В) = С_initial(В) - С_end(В) = 2,5 mol/lit - 1 mol/lit = 1,5 mol/lit.

Answer: υ = 8 ·10^-4mol/lit · s; С(В) = 1,5 mol/lit · s.

11. In the reacting system , the scheme of which

3А + В⇄ 2С + D, the of equilibrium concentration of the substance A, B and C are respectively 0,4, 0,9 and 1,2 mol/lit . Find the initial concentrations of substances A and B.

given:

[А] = 0,4 mol/lit

[В] = 0,9 mol/lit

[С] = 1,2 mol/lit

find:

С(А)_исх.-?

С(В)_исх. -?

Solution:

υ(А)_исх = υ(А)_прор. + υ(А)_разн. = 1,8 mol + 0,4 mol = 2,2 mol

υ(В)_исх = υ(В)_прор. + υ(А)_разн. = 0,6 mol + 0,9 mol = 1,5 mol

Answer: С(А)_исх. = 2,2 mol/lit; С(В)_исх. = mol/lit

12. Calculate the amount of carbon monoxide formed within for 50 seconds , if the rate of the reaction : C_(t) + H₂O_(g) = CO_(g) + H_2(g) of the is 0,3 mol/lit · sec, and the reactor volume is equal 1 liter.

given:

V = 1 lit

τ = 60 sec

ν= 0,3 mol

find:

ν(СО)- ?

Solution:

ν = υ · V · t

ν(СО) = 0,3 mol/lit · sec · 1 lit · 50 sec = 15 mol

Answer: ν(СО)= 15 mol.

13. How to change the speed of the reaction proceeding according to the equation : 2CO(g) + O₂(g) = 2CO₂(g), if:

a) the concentration of oxygen increased in 2 times ?

b) increase the concentration of carbon monoxide in 2 times?

а) given:

[О₂] >in 2 times

find:

υ - ?

Solution:

υ = k[CO]² · [O₂]

1) с₂(СО) = с₁(СО)

2) с₂(О₂) = с₁(О₂) · 2

3) υ = k · с₁(О₂) · 2 = 2 mol/lit·sec.

Answer: When you increase the concentration of oxygen at 2 times the speed of reaction is increased in 2 times.

б) given:

[СО] >in 2 times

find:

υ - ?

Solution:

υ = k[CO]² · [O₂]

1) с₂(СО) = с₁(СО) · 2²

2) с₂(О₂) = с₁(О₂)

3) υ = k · с₁(О₂) · с₁(СО) · 2² = k ·4 = 4 mol/lit · sec.

Answer: When you increase the concentration of carbon monoxide at 4 times the speed of reaction is increased in 4 times.

14. How many times it is necessary to increase the concentration of the hydrogen sulfide, to the reaction rate 2Н₂S_(g) + SО_2(g) = 2S_(s) + 2Н₂О_(g) risen in the 9 times ?

given:

υ>in 9 times

find:

Δс(Н₂S) - ?

Solution:

υ = k·с(Н₂S)² · с(SO₂)

υ = k·с(Н₂S)² · с(SO₂)

υ = 9 mol/lit · с (by condition problem)

1) с₂(Н₂S) = с₁(Н₂S) · х²

2) с₂(SО₂) = с₁(SО₂)

3) 9 mol/lit = k · х²

х = 3 mol/lit·s.

Answer: to increase the rate of this reaction is 9 times necessary to increase the concentration of hydrogen sulfide in 3 times.

15. Write an the expression depending on the speed of the forward and reverse reaction on the concentration reactants of the substances for the following processes :

а) 2SO_3(g) ⇄ 2SO_2(g) + O_2(g)

b) FеO_(кр) + СO_(g) ⇄ Fе _{( кр)} + CO_2(g)

How to change the speed of the forward and reverse reactions,

If the increase pressure system in 3 times ?

given:

find:

Solution:

а) Homogeneous system , then the speed of forward and reverse reaction expressed respectively :

υ₁ = k₁ · [SO₃]² (direct reaction),

υ₂ = k₂ · [SO₂]² · [O₂] (reverse reaction),

With increasing pressure in three times the concentration of substances also increased the by 3 times.

The rate of direct reaction after increasing pressure is expressed by:

υ₁ = k₁ · (3[SO₃]²)

then,

the rate of the forward reaction will increase by 9 times .

For the reverse reaction after increasing pressure 3 times the expression law of mass action will take the form :

υ₂ = k₂ · (3[SO₃]²) · (3[О₂]).

тогда,

rate of the reverse reaction will increase by 27 times .

b) a heterogeneous system , then:

υ₁ = k₁ · [СO] (direct reaction),

υ₂ = k₂ · [СO₂] (reverse reaction)

Consequently,

the rate of the forward reaction will increase by 3 times.

For the reverse reaction :

speed is also increased by 3 times.

Answer: a) the speed of the forward reaction increases by 9 times .

For the reverse reaction the rate increases to 27 times .

b ) the rate of the forward reaction will increase by 3 times.

For the reverse reaction speed is also increased by 3 times.

Basic concepts and stoichiometric laws of chemistry .

1. 5 liters of nitrogen under a pressure of 2 atm , 2 l of oxygen under a pressure of 2.5 atm and 3 liters of carbon dioxide under a pressure of 5 atm of the mixed, the wherein volume of provided mixture equal to 15 liters . Calculate the pressures and is a mixture of the partial pressures of each gas.

given:

V₁(N₂) = 5 lit

P₁ = 2 atm

V₂(О₂) = 2 lit

P₂ = 2,5 atm

V₃(СО₂) = 3 lit

P₃ = 5 atm

V₄(mixture) = 5 lit

find:

P₄ - ? P(N₂) - ?

P(О₂) - ? P(СО₂) - ?

Solution:

From the law of Boyle - Mariotte

P₁ ·V₁ = P₂ ·V₂

P₄ = P(N₂) + P(О₂) + P(СО₂) = 0,67 atm + 0,33 atm + 1,0 atm = 2 atm.

Answer: P₄ = 2 atm,P(N₂) = 0,67 atm, P(О₂) = 0,33 atm, P(СО₂) =1,0 atm.

2. Gas at 15 ° C and a pressure of 760 mm Hg.Art. occupies a volume 2 liters. Bring the volume gas to normal conditions .

given:

P₁ = 750 mm Hg.art.

Т₁ = 15⁰ = 15 + 273 = 288

Т₂ = 273⁰С

P₂ = 760 mm Hg.art.

V₁ = 2 lit

find:

V₂ - ?

Solution:

On the basis of the combined Act defines volume V₂

Answer: V₂ = 1,9 lit.

3. A mixture consisting of 2 moles of hydrogen , some number the of moles of oxygen and 1 mol of nitrogen at 20 ° C and a pressure of 4 atm , occupies a volume of 40 liters. Calculate the number of moles of oxygen in the mixture and the partial pressure of each gas .

given:

ν₁(Н₂) = 2 mol

ν₂(N₂) = 1 mol

P = 4 atm

Т = 20⁰С = 293 К

V(mixture) = 40 lit

find:

ν₃(О₂) - ? P(Н₂) - ?

P(N₂) - ?P(О₂) - ?

Solution:

From equation Mendeleev - Clapeyron total number of moles of all the gases making up the mixture

The number of moles of oxygen in the mixture is equal to:

ν₃(О₂) = ν_mixture - ν₁(Н₂) - ν₂(N₂) = 6,66 mol - 2 mol - 1 mol = 3,66 mol.

The partial pressure each of gas calculated by the equation :

Answer: ν₃(О₂) = 3,66 mol;P(N₂) = 0,6 atm, P(О₂) = 2,2 atm, P(Н₂) =1,2 atm.

4. In a 500 ml flask at 25 ° C is of 0.615 g of nitric oxide (II). Is gas pressure in the flask ?

given:

V = 500 ml = 0,5 lit = 0,5 ·10^-3m³

m = 0,615 g

Т = 20⁰С = 293 K

М(NO) = 30 g/mol

find:

P- ?

Solution:

From equation Mendeleev - Clapeyron find :

Answer:101592,6 Па.