Bu'aan kutaa 12ffaa ba'eera; bu'aa kee eessatti akka ilaaltu

Yaadannoo boqonnaa fi gaaffii shaakalaa

Boqonnaa 5 Keemistirii kutaa 11: Iquiliibriyeemii Keemikaalaa

Wanti fuula kana irra jiru hundi boqonnaa 5 kitaaba Keemistirii kutaa 11 tiif: wanta boqonnaan haguugu, yaadannoo gabaabaa, gaaffii shaakalaa deebii isaanii waliin, fi viidiyoo barnootaa bakka jirutti.

Google Play irraa buufadhaa
3
Kaardii
10
Gaaffii shaakalaa boqonnaa 5
222–263
Fuula kitaabaa boqonnaa 5

Boqonnaan 5 maal haguuga

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

Yaadannoo gabaabaa boqonnaa 5

Kaardiiwwan fuula Keemistirii kutaa 11 irratti boqonnaa kana jala galmaa'an. Tokko tokkoon bani.

Yoom akka itti fayyadamtu

Jijjiiramni dhiibbaa yookiin volume sirna equilibrium gaasii gara kamitti akka siqsu tilmaamuuf fayyadami.

Of eeggadhu

Dhiibbaa dabaluun equilibrium gara cinaa moles gaasii xiqqaatti siqsa. Volume dhaabbataa irratti inert gas dabaluun siqinsa hin fidu.

Grade 11 Chemistry, fuula 222-263 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Yoom akka itti fayyadamtu

Sirna heterogeneous haalota maatarii adda addaa qabaniif ibsa equilibrium constant yeroo barreessitu fayyadami.

Of eeggadhu

Concentrations isaanii dhaabbataa waan ta'eef, solids qulqulluu (s) yookiin liquids qulqulluu (l) ibsa KC yookiin KP keessatti gonkumaa hin galchin.

Grade 11 Chemistry, fuula 222-263 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

KP=KC(RT)ΔnK_P = K_C(RT)^{\Delta n}
KPK_P
equilibrium constant dhiibbaa kutaatiin ibsamuatmΔn\text{atm}^{\Delta n}
KCK_C
equilibrium constant kuusaa moolariitiin ibsamu(mol L−1)Δn(\text{mol}\,\text{L}^{-1})^{\Delta n}
RR
Dhaabbataa gaasii aayidiyaliiL atm K−1 mol−1\text{L}\,\text{atm}\,\text{K}^{-1}\,\text{mol}^{-1}
TT
temperature dhumataaK\text{K}
Δn\Delta n
moles bu'aalee gaasii hir'isuu moles reektantoota gaasii11

Yoom akka itti fayyadamtu

Re'akshinoota gaasii keessatti equilibrium constant kuusaa fi dhiibbaa walitti jijjiiruuf fayyadami.

Of eeggadhu

Delta n yeroo shallagdu gaasota qofa lakkaa'i. Moles bu'aalee fi reektantootaa walqixa yoo ta'an Delta n zeeroo ta'a, KP fi KC walqixa ta'u.

Grade 11 Chemistry, fuula 222-263 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Gaaffii shaakalaa boqonnaa 5

Gaaffii filannoo 10 boqonnaa kana irratti barreeffaman, tokkoon tokkoon isaanii deebii fi sababa qabu. Deebii banuu dura yaali. Gaaffiin kun Ingiliffaa tiin.

Gaaffii 1

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

  1. A.The concentrations of reactants and products are always equal.
  2. B.The rate of the forward reaction is equal to the rate of the reverse reaction.
  3. C.The reaction has stopped completely.
  4. D.Only products are present in the mixture.
Deebii fi sababa agarsiisi

Deebii: B The rate of the forward reaction is equal to the rate of the reverse reaction.

Sababa

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.

Gaaffii 2

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

  1. A.The reaction goes only from left to right.
  2. B.The reaction can go in both the forward and the reverse direction.
  3. C.The reaction has stopped.
  4. D.Only a catalyst is present in the reaction.
Deebii fi sababa agarsiisi

Deebii: B The reaction can go in both the forward and the reverse direction.

Sababa

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.

Gaaffii 3

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

  1. A.K<sub>C</sub> = [CO<sub>2</sub>]
  2. B.K<sub>C</sub> = [CaO][CO<sub>2</sub>] / [CaCO<sub>3</sub>]
  3. C.K<sub>C</sub> = [CaO][CO<sub>2</sub>]
  4. D.K<sub>C</sub> = [CaCO<sub>3</sub>] / [CO<sub>2</sub>]
Deebii fi sababa agarsiisi

Deebii: A K<sub>C</sub> = [CO<sub>2</sub>]

Sababa

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

Gaaffii 4

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

  1. A.10<sup>10</sup>
  2. B.10<sup>-10</sup>
  3. C.1
  4. D.0.5
Deebii fi sababa agarsiisi

Deebii: A 10<sup>10</sup>

Sababa

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

Gaaffii 5

At a certain temperature, KC = 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 QC, and which way will the reaction go to reach equilibrium?

  1. A.Q<sub>C</sub> = 1.67, so the reaction goes forward (to the right).
  2. B.Q<sub>C</sub> = 1.67, so the reaction goes in reverse (to the left).
  3. C.Q<sub>C</sub> = 0.60, so the reaction goes forward (to the right).
  4. D.Q<sub>C</sub> = 13.8, so the mixture is already at equilibrium.
Deebii fi sababa agarsiisi

Deebii: A Q<sub>C</sub> = 1.67, so the reaction goes forward (to the right).

Sababa

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

Gaaffii 6

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

  1. A.0.037 M
  2. B.3.87 M
  3. C.0.0037 M
  4. D.104.4 M
Deebii fi sababa agarsiisi

Deebii: A 0.037 M

Sababa

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

Gaaffii 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?

  1. A.Increase the pressure by decreasing the volume of the vessel.
  2. B.Increase the temperature.
  3. C.Add an inert gas while keeping the volume fixed.
  4. D.Add a catalyst.
Deebii fi sababa agarsiisi

Deebii: A Increase the pressure by decreasing the volume of the vessel.

Sababa

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.

Gaaffii 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?

  1. A.The yield of SO<sub>3</sub> increases.
  2. B.The yield of SO<sub>3</sub> decreases.
  3. C.The yield of SO<sub>3</sub> stays the same.
  4. D.The yield of SO<sub>3</sub> becomes zero.
Deebii fi sababa agarsiisi

Deebii: A The yield of SO<sub>3</sub> increases.

Sababa

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.

Gaaffii 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 KC of the overall reaction?

  1. A.9.1 &times; 10<sup>-16</sup>
  2. B.1.4 &times; 10<sup>-16</sup>
  3. C.7.9 &times; 10<sup>-16</sup>
  4. D.2.4 &times; 10<sup>30</sup>
Deebii fi sababa agarsiisi

Deebii: A 9.1 &times; 10<sup>-16</sup>

Sababa

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

Gaaffii 10

At 430 °C, KC = 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?

  1. A.Q<sub>C</sub> is about 19.4, so the reaction goes forward to form more HI.
  2. B.Q<sub>C</sub> is about 19.4, so the reaction goes in reverse to form more H<sub>2</sub> and I<sub>2</sub>.
  3. C.Q<sub>C</sub> is 54.3, so the mixture is already at equilibrium.
  4. D.Q<sub>C</sub> is about 0.052, so the reaction goes in reverse.
Deebii fi sababa agarsiisi

Deebii: A Q<sub>C</sub> is about 19.4, so the reaction goes forward to form more HI.

Sababa

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

Appiin boqonnaa kana irratti gaaffii 12 qaba; utuu hojjettuu ni sirreessa. Kan asii jiran 10 jalqabaa ti.

Fuula kitaabaa boqonnaa 5

Boqonnaan 5 Maxxansa Ministeera Barnootaa keessatti fuula 222 hanga fuula 263 ti.

Maxxansa Ministeera Barnootaa, Ingiliffaa, Fuula 342

PDF buufadhaa

Yaadannoo gabaabaa Keemistirii kutaa 11 hunda

Boqonnaa kana appii keessatti shaakali

Temari boqonnaa kana gara shaakala yeroon daangeffametti jijjiira; utuu hojjettuu ni sirreessa; wanti si hin galinis barsiisaa AI tiin ni ibsama.

Bilisaan jalqabi

Temari.et appii Android qaba; Google Play irratti bilisaan argama.

Google Play irraa buufadhaaWaa'ee appii Temari
SeeniHerrega kee bani