At a certain temperature, K C = 13.8 for CO(g) + Cl 2 (g) ⇌ COCl 2 (g). A mixture has [CO] = 2.5 mol L -1 , [Cl 2 ] = 1.2 mol L -1 and [COCl 2 ] = 5.0 mol L -1 . What is the reaction quotient Q C , and which way will the reaction go to reach equilibrium?

Q C = 1.67, so the reaction goes forward (to the right).

Q C = 1.67, so the reaction goes in reverse (to the left).

Q C = 0.60, so the reaction goes forward (to the right).

Q C = 13.8, so the mixture is already at equilibrium.

A fertiliser plant in Addis Ababa makes ammonia by the Haber process: N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g). The manager wants more ammonia at equilibrium. Which change, at constant temperature, will increase the yield of NH 3 ?

Add an inert gas while keeping the volume fixed.

At 430 °C, K C = 54.3 for H 2 (g) + I 2 (g) ⇌ 2HI(g). A 3.5 L vessel contains 0.0218 mol H 2 , 0.0145 mol I 2 and 0.0783 mol HI. Which statement is correct?

Q C is about 19.4, so the reaction goes in reverse to form more H 2 and I 2 .

Q C is 54.3, so the mixture is already at equilibrium.

Q C is about 0.052, so the reaction goes in reverse.

የምዕራፍ ማስታወሻዎችና የመልመጃ ጥያቄዎች

የ11ኛ ክፍል ኬሚስትሪ ምዕራፍ 5፦ ኬሚካላዊ ኢኩሊብሪየም

በዚህ ገጽ ያለው ሁሉ ለ11ኛ ክፍል የኬሚስትሪ መፅሐፍ ምዕራፍ 5 ነው፦ ምዕራፉ የሚሸፍነው፣ አጭር ማስታወሻዎቹ፣ የመልመጃ ጥያቄዎች ከመልሶቻቸው ጋር፣ እና ቪዲዮ ባለበት የትምህርቱ ቪዲዮ።

3
ካርዶች: 10
የምዕራፍ 5 የመልመጃ ጥያቄዎች: 222–263
የምዕራፍ 5 የመፅሐፍ ገጾች

ምዕራፍ 5 የሚሸፍነው

1.Reversible and Irreversible Reactions
2.Attainment and Characteristics of Chemical Equilibria
3.Conditions for Attainment of Chemical Equilibria
4.Equilibrium Expression and Equilibrium Constant
5.Applications of Equilibrium Constant
6.Changing Equilibrium Conditions: Le-Chatelier's Principle
7.Equilibrium and Industry

የምዕራፍ 5 አጭር ማስታወሻዎች

በ11ኛ ክፍል የኬሚስትሪ ገጽ ላይ በዚህ ምዕራፍ ሥር የተመዘገቡት ካርዶች። አንድ በአንድ ይክፈቱ።

የLe Chatelier መርህ እና የግፊት ለውጥሕግ

መቼ እንደሚጠቀሙበት

የግፊት ወይም የvolume ለውጥ የጋዝ equilibrium ሥርዓትን ወደ የትኛው አቅጣጫ እንደሚያሸጋግረው ለመተንበይ ይጠቀሙ።

ይጠንቀቁ

ግፊት ሲጨምር equilibrium አነስተኛ የጋዝ moles ወዳለበት አቅጣጫ ይሸጋገራል። በቋሚ volume ላይ inert gas መጨመር ምንም ሽግግር አያመጣም።

ከGrade 11 Chemistry፣ ገጽ 222-263 የተዘጋጀ፤ ከመውጣቱ በፊት ሁለት ጊዜ ተረጋግጧል

ንጹሕ Solids እና Liquids በEquilibrium ቀመር ውስጥ ማካተትየተለመደ ስህተት

መቼ እንደሚጠቀሙበት

የተለያዩ የቁስ አካል ሁኔታዎችን ለያዙ heterogeneous systems የequilibrium constant ቀመር ሲጻፍ ይጠቀሙ።

ይጠንቀቁ

ንጹሕ solids (s) ወይም ንጹሕ liquids (l) concentrations ቋሚ ስለሆኑ በKC ወይም በKP ቀመር ውስጥ በፍጹም እንዳይካተቱ።

ከGrade 11 Chemistry፣ ገጽ 222-263 የተዘጋጀ፤ ከመውጣቱ በፊት ሁለት ጊዜ ተረጋግጧል

በKP እና በKC መካከል ያለው ዝምድናቀመር

K_P = K_C(RT)^{\Delta n}

K_P = K_C(RT)^{\Delta n}

$K_P$: በከፊል ግፊት የሚገለጽ equilibrium constant $\text{atm}^{\Delta n}$
$K_C$: በሞላር ክምችት የሚገለጽ equilibrium constant $(\text{mol}\,\text{L}^{-1})^{\Delta n}$
$R$: የአይዲያል ጋዝ ቋሚ $\text{L}\,\text{atm}\,\text{K}^{-1}\,\text{mol}^{-1}$
$T$: ፍጹም temperature $\text{K}$
$\Delta n$: የጋዝ ውጤቶች moles ሲቀነስ የጋዝ መነሻዎች moles $1$

መቼ እንደሚጠቀሙበት

ለጋዝ phase reactions በconcentration እና በpressure equilibrium constant መካከል ለመቀየር ይጠቀሙ።

ይጠንቀቁ

Delta n ሲሰላ የጋዝ ዝርያዎችን ብቻ ይቁጠሩ። የጋዝ products እና reactants moles እኩል ሲሆኑ Delta n ዜሮ ይሆናል፤ KP እና KC እኩል ይሆናሉ።

ከGrade 11 Chemistry፣ ገጽ 222-263 የተዘጋጀ፤ ከመውጣቱ በፊት ሁለት ጊዜ ተረጋግጧል

የምዕራፍ 5 የመልመጃ ጥያቄዎች

በዚህ ምዕራፍ ላይ የተጻፉ 10 የምርጫ ጥያቄዎች፣ እያንዳንዳቸው መልሱና ምክንያቱ አላቸው። መልሱን ከመክፈትዎ በፊት ይሞክሩ። ጥያቄዎቹ በእንግሊዝኛ ናቸው።

ጥያቄ 1

Which statement is correct for a reaction that has reached equilibrium?

A.The concentrations of reactants and products are always equal.
B.The rate of the forward reaction is equal to the rate of the reverse reaction.
C.The reaction has stopped completely.
D.Only products are present in the mixture.

መልሱንና ምክንያቱን ያሳዩ

መልስ፦ B The rate of the forward reaction is equal to the rate of the reverse reaction.

ምክንያቱ

At equilibrium the forward and reverse reactions still happen, but they happen at the same rate. The concentrations stay constant. They are not necessarily equal, so option A is wrong, and the reaction has not stopped, so option C is wrong.

ጥያቄ 2

In a chemical equation, what does a double arrow (⇌) show?

A.The reaction goes only from left to right.
B.The reaction can go in both the forward and the reverse direction.
C.The reaction has stopped.
D.Only a catalyst is present in the reaction.

መልሱንና ምክንያቱን ያሳዩ

መልስ፦ B The reaction can go in both the forward and the reverse direction.

ምክንያቱ

A double arrow means the reaction is reversible. Reactants form products (forward reaction), and products form reactants again (reverse reaction). A single arrow is used for irreversible reactions that go in one direction only.

ጥያቄ 3

For the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g), which expression is the correct K_C?

A.KC = [CO2]
B.KC = [CaO][CO2] / [CaCO3]
C.KC = [CaO][CO2]
D.KC = [CaCO3] / [CO2]

መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A KC = [CO2]

ምክንያቱ

Pure solids and pure liquids are not included in the K_C expression. CaCO₃ and CaO are both solids, so only the gas CO₂ appears in the expression.

ጥያቄ 4

A reaction goes essentially to completion if its equilibrium constant is greater than what value?

A.1010
B.10-10
C.1
D.0.5

መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A 1010

ምክንያቱ

The chapter says a reaction goes essentially to completion if K_C or K_P is greater than 10¹⁰. It does not go at all if the value is less than 10^-10.

ጥያቄ 5

At a certain temperature, K_C = 13.8 for CO(g) + Cl₂(g) ⇌ COCl₂(g). A mixture has [CO] = 2.5 mol L^-1, [Cl₂] = 1.2 mol L^-1 and [COCl₂] = 5.0 mol L^-1. What is the reaction quotient Q_C, and which way will the reaction go to reach equilibrium?

A.QC = 1.67, so the reaction goes forward (to the right).
B.QC = 1.67, so the reaction goes in reverse (to the left).
C.QC = 0.60, so the reaction goes forward (to the right).
D.QC = 13.8, so the mixture is already at equilibrium.

መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A QC = 1.67, so the reaction goes forward (to the right).

ምክንያቱ

Q_C = [COCl₂] / ([CO][Cl₂]) = 5.0 / (2.5 × 1.2) = 1.67. Since Q_C is less than K_C (13.8), the reaction must go forward to make more product and reach equilibrium.

ጥያቄ 6

For the decomposition of CaCO₃ at 1273 K, the pressure of CO₂ in equilibrium is K_P = 3.87 atm. Use R = 0.082 L atm K^-1 mol^-1. Which value is closest to K_C? (Use K_C = K_P / RT, because the change in moles of gas is +1.)

A.0.037 M
B.3.87 M
C.0.0037 M
D.104.4 M

መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A 0.037 M

ምክንያቱ

RT = 0.082 × 1273 = 104.4. Then K_C = 3.87 / 104.4 = 0.037 M. Dividing by RT is the right step here because K_P = K_C(RT)^Δn with Δn = +1.

ጥያቄ 7

A fertiliser plant in Addis Ababa makes ammonia by the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The manager wants more ammonia at equilibrium. Which change, at constant temperature, will increase the yield of NH₃?

A.Increase the pressure by decreasing the volume of the vessel.
B.Increase the temperature.
C.Add an inert gas while keeping the volume fixed.
D.Add a catalyst.

መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A Increase the pressure by decreasing the volume of the vessel.

ምክንያቱ

There are 4 mol of gas on the left and 2 mol on the right, so the forward reaction makes fewer gas molecules. Increasing the pressure shifts the equilibrium toward fewer gas molecules, which is the forward direction. A catalyst speeds up both reactions but does not change the yield.

ጥያቄ 8

In the Contact process, the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is exothermic. If the temperature at equilibrium is lowered, what happens to the yield of SO₃ according to Le Chatelier's principle?

A.The yield of SO3 increases.
B.The yield of SO3 decreases.
C.The yield of SO3 stays the same.
D.The yield of SO3 becomes zero.

መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A The yield of SO3 increases.

ምክንያቱ

Because the forward reaction is exothermic, it gives out heat. Lowering the temperature removes heat, so the system shifts in the direction that makes heat, which is the forward direction. More SO₃ is made. In real plants a compromise temperature of about 450 °C is used, because lower temperatures make the reaction too slow.

ጥያቄ 9

Reactions can be added to give an overall reaction. For ½N₂(g) + ½O₂(g) + ½Br₂(g) ⇌ NOBr(g), the equilibrium constant of NO(g) + ½Br₂(g) ⇌ NOBr(g) is K₂ = 1.4, and the equilibrium constant of ½N₂(g) + ½O₂(g) ⇌ NO(g) is K₁' = 6.5 × 10^-16. What is the K_C of the overall reaction?

A.9.1 × 10-16
B.1.4 × 10-16
C.7.9 × 10-16
D.2.4 × 1030

መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A 9.1 × 10-16

ምክንያቱ

When reactions are added, the overall K is the product of the individual K values, not the sum. So K_overall = K₂ × K₁' = 1.4 × 6.5 × 10^-16 = 9.1 × 10^-16. Adding the numbers (option C) is a common mistake.

ጥያቄ 10

At 430 °C, K_C = 54.3 for H₂(g) + I₂(g) ⇌ 2HI(g). A 3.5 L vessel contains 0.0218 mol H₂, 0.0145 mol I₂ and 0.0783 mol HI. Which statement is correct?

A.QC is about 19.4, so the reaction goes forward to form more HI.
B.QC is about 19.4, so the reaction goes in reverse to form more H2 and I2.
C.QC is 54.3, so the mixture is already at equilibrium.
D.QC is about 0.052, so the reaction goes in reverse.

መልሱንና ምክንያቱን ያሳዩ

መልስ፦ A QC is about 19.4, so the reaction goes forward to form more HI.

ምክንያቱ

First find the concentrations by dividing by 3.5 L: [H₂] = 0.00623, [I₂] = 0.00414, [HI] = 0.02237 mol L^-1. Then Q_C = (0.02237)² / (0.00623 × 0.00414) = about 19.4. Since Q_C is less than K_C (54.3), the reaction goes forward.

መተግበሪያው በዚህ ምዕራፍ ላይ 12 ጥያቄዎች አሉት፤ እየሠሩ ያርምዎታል። እዚህ ያሉት የመጀመሪያዎቹ 10 ናቸው።

የምዕራፍ 5 የመፅሐፍ ገጾች

ምዕራፍ 5 በየትምህርት ሚኒስቴር እትም ከገጽ 222 እስከ ገጽ 263 ይደርሳል።

የትምህርት ሚኒስቴር እትም, እንግሊዝኛ, 342 ገጾች

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ሁሉም የ11ኛ ክፍል ኬሚስትሪ አጭር ማስታወሻዎች

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