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+ I 2 (g)

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+ I 2 (g)
(1) Consider the following reactions and their associated equilibrium
constants:
For the reaction A + 2B
A. Kc = K1 + K2
D. Kc = (K1)(K2)
D + E, having equilibrium constant Kc,
B. Kc = K1/K2
C. Kc = K1 - K2
E. Kc = K2/K1
(2) Consider the following equilibria:
Calculate the equilibrium constant for the reaction
SO2(g) + NO3(g)
A. 78
D. 3.2  10-10
SO3(g) + NO2(g).
B. 1.3  10-2
E. 6.1  103
C. 1.6  10-4
‫ العالقة بين الكيمياء الحركية واالتزان الكيميائي‬8.3
Chemical Kinetics and Chemical Equilibrium
A + 2B
kf
kr
ratef = kf [A][B]2
AB2
rater = kr [AB2]
Equilibrium
ratef = rater
kf [A][B]2 = kr [AB2]
kf
[AB2]
= Kc =
kr
[A][B]2
‫ ما هو مدلول ثابت اإلتزان‬8.4
What Does the Equilibrium constant tell us
The reaction quotient (Qc) is calculated by substituting the
initial concentrations of the reactants and products into the
equilibrium constant (Kc) expression. IF
•
Qc > Kc system proceeds from right to left to reach equilibrium
•
Qc = Kc the system is at equilibrium
•
Qc < Kc system proceeds from left to right to reach equilibrium
‫قيمة القسمة ‪(Quotient‬يرمز لها بالرمز ‪ )Q‬هي حاصل ضرب تركيز النواتج على حاصل ضرب تركيز‬
‫المتفاعالت في لحظة معينة ‪ ،‬ويتم حساب قيمها ألي تفاعل عكوس باستعمال نفس قانون االتزان‪:‬‬
‫‪aA + bB‬‬
‫‪cC + dD‬‬
‫‪[C]C [D]d‬‬
‫‪Q ‬‬
‫‪[A]a .[B]b‬‬
‫ال يشترط أن تكون تراكيز المتفاعالت والنواتج في لحظة مساوية لتراكيزها عند االتزان عندما نقوم بإيجاد قيمة القسمة‬
‫‪Q‬حيث عندما تكون قيمة ‪ = Q‬صفر هذا يدل على أن تركيز النواتج صفر ‪،‬‬
‫أما عندما تكون قيمة ‪Q‬أقل من قيمة ثابت االتزان ‪ KC‬فهذا يدلنا على أن التفاعل ‪:‬‬
‫(‪ )1‬لم يصل لحالة االتزان‬
‫(‪ )2‬ويجب أن نتحكم في التفاعل لكي يتجه ناحية اليمين لتكوين نواتج‪.‬‬
‫أما عندما تكون قيمة ‪ Q‬أكبر من قيمة ثابت االتزان ‪KC‬فنعرف أن التفاعل ‪:‬‬
‫(‪ )1‬لم يصل لحالة االتزان‬
‫(‪ )2‬ويجب أن نتحكم في التفاعل لكي يتجه ناحية اليسار لتكوين كمية أكبر‬
‫من المتفاعالت للوصول لحالة االتزان‪.‬‬
‫أما إذا وجدت قيمة ‪ = Q‬قيمة ‪ ، KC‬فنعرف أن تراكيز المتفاعالت والنواتج أصبحت مساوية للتراكيز‬
‫عند حالة االتزان الكيميائي للتفاعل العكوس‪.‬‬
‫وأهمية معرفة قيمة القسمة ‪ Q‬تعطينا فكرة عن مسار التفاعل العكوس وكيفية التعامل مع التفاعل للوصول لحالة االتزان‬
‫أما عندما تكون قيمة ‪ Q‬أقل من قيمة ثابت االتزان ‪ Kc‬فهذا يدلنا على أن التفاعل‪:‬‬
‫(‪ )1‬لم يصل لحالة االتزان‬
‫(‪ )2‬ويجب أن نتحكم في التفاعل لكي يتجه ناحية اليمين لتكوين نواتج‪.‬‬
‫أما عندما تكون قيمة ‪ Q‬أكبر من قيمة ثابت االتزان ‪ Kc‬فنعرف أن التفاعل ‪:‬‬
‫(‪ )1‬لم يصل لحالة االتزان‬
‫(‪ )2‬ويجب أن نتحكم في التفاعل لكي يتجه ناحية اليسار لتكوين كمية أكبر‬
‫من المتفاعالت للوصول لحالة االتزان‪.‬‬
‫أما إذا وجدت قيمة ‪ = Q‬قيمة ‪: Kc‬‬
‫فنعرف أن تراكيز المتفاعالت والنواتج أصبحت مساوية للتراكيز عند حالة االتزان‬
‫الكيميائي للتفاعل العكوس‪.‬‬
‫تزداد قيم الـ ‪Q‬‬
‫‪Q<K‬‬
‫متفاعالت فقط‬
‫يتجه التفاعل جهة اليسار لتكوين المتفاعالت‬
‫جهة السهم‬
‫‪Kc‬‬
‫‪Q>K‬‬
‫متفاعالت فقط‬
‫يتجه التفاعل جهة اليمين لتكوين نواتج‬
‫يتجه التفاعل جهة النواتج عندما تكون قيمة ‪ Q‬أصغر من ‪K‬‬
‫يتجه التفاعل جهة المتفاعالت عندما تكون قيمة ‪ Q‬أكبر من ‪K‬‬
‫ويكو ن التفاعل وصل لحالة االتزان عندما ‪K = Q‬‬
(1) At 700 K, the reaction 2SO2(g) + O2(g)
2SO3(g) has the
equilibrium constant Kc = 4.3  106, and the following concentrations
are present: [SO2] = 0.10 M; [SO3] = 10. M; [O2] = 0.10 M. Is the mixture
at equilibrium? If not at equilibrium, in which direction (as the
equation is written), left to right or right to left, will the reaction
proceed to reach equilibrium?
A. Yes, the mixture is at equilibrium.
B. No, left to right
C. No, right to left
D. There is not enough information to be able to predict the direction.
(2) At 700 K, the reaction 2SO2(g) + O2(g)
2SO3(g) has the
equilibrium constant Kc = 4.3  106, and the following concentrations
are present: [SO2] = 0.010 M; [SO3] = 10.M; [O2] = 0.010 M. Is the
mixture at equilibrium? If not at equilibrium, in which direction (as
the equation is written), left to right or right to left, will the reaction
proceed to reach equilibrium?
A. Yes, the mixture is at equilibrium.
B. No, left to right
C. No, right to left
D. There is not enough information to be able to predict the direction.
(3) For the reaction H2(g) + I2(g)
2HI(g), Kc = 50.2 at 445ºC. If [H2] =
[I2] = [HI] = 1.75  10-3 M at 445ºC, which one of these statements is
true?
A. The system is at equilibrium, thus no concentration changes will occur.
B. The concentrations of HI and I2 will increase as the system approaches
equilibrium.
C. The concentration of HI will increase as the system approaches
equilibrium.
D. The concentrations of H2 and HI will fall as the system moves toward
equilibrium.
(4) For the reaction PCl3(g) + Cl2(g)
PCl5(g) at a particular
temperature, Kc = 24.3. Suppose a system at that temperature is
prepared with [PCl3] = 0.10 M, [Cl2] = 0.15 M, and [PCl5] = 0.60 M.
Which of these statements is true?
A. The reaction is at equilibrium.
B. The reaction will proceed in the direction of forming more PCl5 until
equilibrium is reached.
C. The reaction will proceed in the direction of forming more PCl3 and
Cl2 until equilibrium is reached.
D. None of these statements is true.
Calculating Equilibrium Concentrations
1. Express the equilibrium concentrations of all species in
terms of the initial concentrations and a single unknown x,
which represents the change in concentration.
2. Write the equilibrium constant expression in terms of the
equilibrium concentrations. Knowing the value of the
equilibrium constant, solve for x.
3. Having solved for x, calculate the equilibrium
concentrations of all species.
At 12800C the equilibrium constant (Kc) for the reaction
Br2 (g)
2Br (g)
Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063
M and [Br] = 0.012 M, calculate the concentrations of these
species at equilibrium.
Let x be the change in concentration of Br2
Initial (M)
Change (M)
Equilibrium (M)
[Br]2
Kc =
[Br2]
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
Solve for x
(0.012 + 2x)2
= 1.1 x 10-3
Kc =
0.063 - x
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
-b ± b2 – 4ac
2
x=
ax + bx + c =0
2a
x = -0.0105 x = -0.00178
Initial (M)
Change (M)
Equilibrium (M)
Br2 (g)
2Br (g)
0.063
0.012
-x
+2x
0.063 - x
0.012 + 2x
At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M
At equilibrium, [Br2] = 0.062 – x = 0.0648 M
(1) For the reaction 2NOCl(g)
2NO(g) + Cl2(g), Kc = 8.0 at a certain
temperature. What concentration of NOCl must be put into an empty 4.00 L
reaction vessel in order that the equilibrium concentration of NOCl be
1.00M?
A. 1.26 M
B. 2.25 M
C. 2.50 M
D. 3.52 M
(2) A quantity of liquid methanol, CH3OH, is introduced into a rigid 3.00-L vessel,
the vessel is sealed, and the temperature is raised to 500K. At this
temperature, the methanol vaporizes and decomposes according to the
reaction:
CH3OH(g)
CO(g) + 2 H2(g), Kc= 6.90 10-2.
If the concentration of H2 in the equilibrium mixture is 0.426 M, what mass of
methanol was initially introduced into the vessel?
A. 147 g
B. 74.3 g
C. 33.9 g
D. 49.0 g
E. 24.8 g
(3) When the reaction 2H2S(g)
2H2(g) + S2(g) is carried out at 1065C, Kp =
0.012. Starting with pure H2S at 1065C, what must the initial pressure of H2S
be if the equilibrated mixture at this temperature is to contain 0.250 atm of
H2(g)?
A. 1.06 atm
B. 1.86 atm C. 0.94 atm
D. 0.90 atm
E. 1.52 atm
(4) If the reaction 2H2S(g)
2H2(g) + S2(g) is carried out at 1065C,
Kp = 0.0120. Starting from pure H2S introduced into an evacuated vessel
at 1065C, what will the total pressure in the vessel be at equilibrium if the
equilibrated mixture contains 0.300 atm of H2(g)?
A. 1.06 atm
B. 1.36 atm
C.2.39 atm
D. 4.20 atm
E. 1.51 atm
(5) Sodium carbonate, Na2CO3(s), can be prepared by heating sodium
bicarbonate, NaHCO3(s).
2NaHCO3(s)
Na2CO3(s) + CO2(g) + H2O(g) Kp = 0.23 at 100ºC
If a sample of NaHCO3 is placed in an evacuated flask and allowed to
achieve equilibrium at 100ºC, what will the total gas pressure be?
A. 0.46 atm
B. 0.96 atm
C. 0.23 atm
D. 0.48 atm
E. 0.11 atm
(6) Hydrogen iodide decomposes according to the equation:
2HI(g)
H2(g) + I2(g), for which Kc = 0.0156 at 400ºC. 0.550 mol HI was
injected into a 2.00 L reaction vessel at 400ºC. Calculate the concentration
of H2 at equilibrium.
A. 0.275 M
B. 0.138 M
C. 0.0275 M D. 0.0550 M E. 0.220 M
(7) 75.0 g of PCl5(g) is introduced into an evacuated 3.00 L vessel and allowed
to reach equilibrium at 250ºC. PCl5(g)
PCl3(g) + Cl2(g). If Kp = 1.80 for
this reaction, what is the total pressure inside the vessel at equilibrium?
A. 2.88 atm B. 2.27 atm
C. 4.54 atm
D. 7.44 atm
E. 9.69 atm
:‫) كالتالي‬7( ‫حل تمرين‬
Volume 3L
T= 25 + 273 = 523 K
Kp = 1.80
M.wt.
208.25g/mol 137.35g/mol 70.9 g/mol
Wt.(gm)
75 gm
? gm
? gm
at equil.
PCl5(g)
PCl3(g) + Cl2(g)
Initial
[0.12]
Zero
Zero
At equil [0.12– x]
[x]
[x]
___________________________________________________
Kp=1.80 = Kc (RT)Dn=2-1=1
Kc = Kp / RT = 1.8/0.0821x 523 = 0.0419
Kc = [x][x] / [0.12-x] = 0.0419
x = 0.053
Kc = [0.053] [0.053] / [0.12–0.053] = 0.0419
___________________________________________
At equil
PCl5(g)
PCl3(g)
At equil
[0.12–0.053]
[0.053]
+
Molarity
[0.067] M
[0.053] M
Mole
[0.067]x3= 0.201mol
[0.053]x3= 0.159mol
Grams
41.86 gm
+
21.84 gm +
PPCl5 x 3L = 0.201 x 0.0821 x 523
PPCl3 x 3L = 0.159 x 0.0821 x 523
PCl2 x 3L = 0.159 x 0.0821 x 523
Cl2(g)
[0.053]
[0.053] M
[0.053]x3= 0.159
11.27 gm = 75 gm
PPCl5 = 2.88
PPCl3 = 2.28
PCl2 = 2.28
PTotal = PPCl5 + PPCl3 + PCl2 = 2.88 + 2.28 + 2.28 = 7.44 atm
‫لتأكيد الحل‬
‫ العوامل المؤثرة على اإلتزان الكيميائي‬8.5
Factors that Affect chemical Equilibrium
Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the
system adjusts in such a way that the stress is partially offset
as the system reaches a new equilibrium position.
• Changes in Concentration
N2 (g) + 3H2 (g)
2NH3 (g)
Equilibrium
shifts left to
offset stress
Add
NH3
Le Châtelier’s Principle
(1) Changes in Concentration continued
‫التغيير في التركيز باستمرار‬
Remove
Add
Remove
Add
aA + bB
cC + dD
Change
‫التغيير الحادث في االتزان‬
Shifts the Equilibrium
‫إزاحة االتزان جهة‬
a) Increase concentration of product(s)
b) Decrease concentration of product(s)
c) Increase concentration of reactant(s)
d) Decrease concentration of reactant(s)
right
left
left
right
Le Châtelier’s Principle
(2) Changes in Volume
(3) Changes in Pressure
A (g) + B (g)
C (g)
Change
Shifts the Equilibrium
Increase pressure
Decrease pressure
Increase volume
Decrease volume
Side with fewest moles of gas
Side with most moles of gas
Side with most moles of gas
Side with fewest moles of gas
‫تدريب‪:‬‬
‫)‪ (1‬زيادة ضغط التفاعل المتزن التالي‪:‬‬
‫تؤدي إلي‪:‬‬
‫)‪2PH3(g‬‬
‫)‪2P(s) + 3H29g‬‬
‫أ ـ زيادة تركيز ‪ PH3‬ب _ ال تؤثر على تركيز النواتج‬
‫د ـ زيادة تركيز المتفاعالت‬
‫ج ـ زيادة قيمة ‪Kc‬‬
‫(‪ )2‬وجود قوى التجاذب بين جزيئات الغاز تؤدى إلى‪:‬‬
‫ب ‪ -‬زيادة الضغط‬
‫أ ـ نقص الضغط‬
‫د ـ زيادة السرعة الجزيئية‬
‫ج ـ زيادة الحجم‬
‫(‪ )3‬زيادة الضغط على التفاعل ‪2NH3(g) :‬‬
‫عند االتزان يؤدي إلى تقليل الحجم مما يترتب عليه‪:‬‬
‫ب ‪ -‬نقص قيمة ‪ Kc‬العددية‬
‫أ ـ نقص في كميات النواتج‬
‫ج ـ زيادة كمية النواتج د ـ زيادة قيمة ‪ Kc‬العددية‬
‫)‪3H2(g) + N2(g‬‬
‫للتأكد توجد اإلجابة في كتاب مسائل وحلول بالباب الثالث الفصل ‪. 6‬‬
Le Châtelier’s Principle
(4) Changes in Temperature
Change
Increase temperature
Decrease temperature
Exothermic Rx
Endothermic Rx
K decreases
K increases
K increases
K decreases
colder
hotter
‫(‪ )1‬تأثير التركيز على االتزان‪ :‬إزاحة االتزان جهة النواتج ( قيمة ‪ K‬تزداد) عندما يتم‬
‫سحب كمية من المتفاعالت والعكس صحيح‪.‬‬
‫(‪ )2‬تأثير الضغط على االتزان‪ :‬إزاحة االتزان جهة عدد الموالت األقل عند زيادة ضغط‬
‫إناء التفاعل المغلق والعكس صحيح‪.‬‬
‫(‪ )3‬تأثير الحجــم على االتزان‪ :‬إزاحة االتزان جهة عدد الموالت األكبر عند زيادة حجم‬
‫إناء التفاعل المغلق والعكس صحيح‪.‬‬
‫(‪ )4‬تأثير الحرارة على االتزان‪ :‬إزاحة االتزان جهة المتفاعالت (أي قيمة ‪ K‬تزداد) في حالة‬
‫زيادة درجة حرارة نظام التفاعل الطارد للحرارة‬
‫‪ DH0 = - Ve kJ/mol‬والعكس صحيح بالنسبة للتفاعل‬
‫الماص للحرارة ‪ DH0 = + Ve kJ/mol‬عند زيادة الحرارة‪.‬‬
‫(‪ )5‬تأثير الحفز على األتزان ‪ :‬ال يوجد أي تأثير على االتزان من إضافة حافز ولكن‬
‫معدل سرعة الوصول لحالة االتزان تداد في وجود حافز‪.‬‬
Le Châtelier’s Principle
• Adding a Catalyst
• does not change K
• does not shift the position of an equilibrium system
• system will reach equilibrium sooner
uncatalyzed
catalyzed
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.
Chemistry In Action: The Haber Process
N2 (g) + 3H2 (g)
2NH3 (g) DH0 = -92.6 kJ/mol
Le Châtelier’s Principle
Change
Shift Equilibrium
Change Equilibrium
Constant
Concentration
yes
no
Pressure
yes
no
Volume
yes
no
Temperature
yes
yes
Catalyst
no
no
:‫ تتوقف قيمة ثابت االتزان على‬: ‫تدريب‬
‫أ ـ الضغط داخل الوعاء وحجم الوعاء‬
‫ب _ ثبات درجة ح اررة الوعاء‬
‫ج ـ التركيز االبتدائي للمواد المتفاعلة والناتجة من التفاعل د ـ‬
‫جميع اإلجابات صحيحة‬
. 6 ‫للتأكد توجد اإلجابة في كتاب مسائل وحلول بالباب الثالث الفصل‬
(1) For the common allotropes of carbon (graphite and diamond),
C(gr)
C(dia) with equilibrium constant K = 0.32. The molar volumes of
graphite and diamond are, respectively, 5.30 cm3/mol and 3.42 cm3/mol;
DHf of diamond is 1.90 kJ/mol. These data suggest that the formation of
diamond is favored at
A. low temperatures and low pressures. B. high temperatures and low pressures.
C. low temperatures and high pressures. D. high temperatures and high pressures.
(2) Which of these situations will result if some CH4(g) is removed from the
reaction CO(g) + 3H2(g)
CH4(g) + H2O(g) at equilibrium?
A. H2O will be consumed.
B. More CH4 and H2O will be produced.
C. Kp will decrease.
D. More CO will be produced.
E.No change will occur.
(3) In which of these gas-phase equilibria is the yield of products increased
by increasing the total pressure on the reaction mixture?
A. CO(g) + H2O(g)
CO2(g) + H2(g)
B. 2NO(g) + Cl2(g)
C. 2SO3(g)
D. PCl5(g)
2NOCl(g)
2SO2(g) + O2(g)
PCl3(g) + Cl2(g)
(4) The reaction 2NO(g)
N2(g) + O2(g) is exothermic, DHº rxn = -180
kJ/mol. Which one of these statements is true?
A. Kp at 1,000 K is less than Kp at 2,000 K.
B. Kp at 1,000 K is larger than Kp at 2,000 K.
C. The Kp's at 1000 K and 2000 K are the same.
D. Kp depends on total pressure as well as temperature.
(5) When the substances in the equation below are at equilibrium, at P an T,
the equilibrium can be shifted to favor the products by
CuO(s) + H2(g)
H2O(g) + Cu(s) DHº rxn = -2.0 kJ/mol
A. increasing the pressure by means of a moving piston at constant T.
B. increasing the pressure by adding an inert gas such as nitrogen.
C. decreasing the temperature.
D. allowing some gases to escape at constant P and T.
E. adding a catalyst.
(5) The equilibrium constants (expressed in atm) for the chemical reaction
N2(g) + O2(g)
2NO(g) are Kp = 1.1  10-3 and 3.6  10-3 at 2,200 K and
2,500 K, respectively. Which one of these statements is true?
A. The reaction is exothermic, DHº < 0.
B. The partial pressure of NO(g) is less at 2,200 K than at 2,500 K.
C. Kp is less than Kc by a factor of (RT).
D. The total pressure at 2,200 K is the same as at 2,500 K.
E. Higher total pressure shifts the equilibrium to the left.
(6) Consider this reaction at equilibrium:
2SO2(g) + O2(g)
2SO3(g), DHº rxn = -198 kJ/mol
If the volume of the system is compressed at constant temperature, what
change will occur in the position of the equilibrium?
A. a shift to produce more SO2
B. a shift to produce more O2
C. no change
D. a shift to produce more SO3
(7) The reaction 2SO3(g)
2SO2(g) + O2(g) is endothermic. If the
temperature is increased,
A. more SO3 will be produced. B. Kc will decrease. C. Kc will increase.
D. no change will occur in Kc .
E. the pressure will decrease.
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