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Reversible Reactions
Q.1
What is meant by the state of chemical equilibrium?
Correct Verbatim Answer
It is the state in a reversible direction when the rates of forward and reverse become and the reactions equal concentrations of reactants and products remain constant.
Q.2
Define reversible reaction. Give an example.
Correct Verbatim Answer
A reversible reaction is one that both forward and reverse proceeds in directions. Example N, + 3H2(g) = 2NH 3(8)
Equilibrium Constant
Q.3
The change of volume disturbs the equilibrium positions for some of the gas phase reactions but not the equilibrium constant. Why?
Correct Verbatim Answer
Volume change and equilibrium: Change in volume affects the equilibrium position, not the equilibrium constant, which only changes with temperature. Mention the characteristics of 0.4 chemical equilibrium. Characteristics Ans. of chemical equilibrium: • Forward and reverse reaction rates are equal • Concentrations remain constant • Can be reached from either side • Dynamic in nature the attain
Reversible reaction
Law of Mass Action
Q.3
(a) Define and explain the law of mass action and derive the expression for the equilibrium constant.
Correct Verbatim Answer
See Q.6 from theory. (b) Write the expressions for Kc for the following reactions. = Sn*+ Sn?+ (i) (aq) (aq) +2Fe2+ (aq) + 2Fę3+ (ii) Ag t (ag) + Ag(s) (aq) = Fe3+ (ag) + Fe2+ (iii) = 2NO (g) N2(g) + 02(g) = (iv) 4N3(8) + 502(g) = 4N0 (g) +6H,0(g) (V) Ans. (i) Sn?+ (ag) = Sn*t (aq) + 2Fe?+ (aq) + 2Fe3+ [Sn* ][Fe+2 Kc [Sn'*][Fe+? *(ag) + Ag(s) (i) Ag' (ag) + Fe?+ (aq). → Fe3+ (iii) N2(g) +02(g) = 2NO(g) (iv) 4H3(g) +502(g) = 4NO (g) + 6H, 0(g) [NO1[H,01 = (V) PCl5(g) PC/3(g) +C12(g) K=PC, IC,1
Q.4
Write down the Kc for the following reactions. Suppose that the reaction mixture in all the case is 'V' dm (i) CH COOH+CH CHOH - CH,COC, H, + H2O 11) 2H - = H2+12 (ii) N, + 3H, = 2NH,
position of equilibrium which is dynamic in nature and not static. Explain it.
Correct Verbatim Answer
Because reactants and products continuously convert into each other at equal rates; the system is active even though no visible change occurs. It is macroscopically dynamic in nature.
Q.5
In the equilibrium - PC 3(g) + C-2(g) PC15(g) F What is the effect on the following changes? Explain your
Correct Verbatim Answer
wer. (i) if temperature is increased (iii) catalyst is added Ans. The following changes occurs when: (i) Temperature increased • The forward dissociation (PCls→PCl3+Cl2) is endothermic. • Increasing temperature shifts equilibrium to the right (more PCl3, Clz). • Kc increase, because increasing temperature favors the endothermic reaction for forward direction. (ii) Volume of container decreased (pressure increased) • Moles of gases are 1 mole at reactant side, 2 moles at product side. • Higher pressure or lower volume shifts equilibrium to the left (towards fewer moles or product side, forming more PCl5. • After achieving new equilibrium, Ko remains unchanged. (iii) Catalyst added • It speeds up the reaction approach to equilibrium without changing the equilibrium position. • No change in equilibrium composition, also Kc remains unchanged (iv) Chlorine (Cl2) added • Adding Clz a product shifts equilibrium to the left to consume some Clz, forming more PCl5. • Kc remains unchanged, by the addition of Clz.
Q.6
Why do the rates of forward reactions slow down when a reversible reaction approaches the equilibrium stage?
Correct Verbatim Answer
Because the concentration of reactants decreases, lowering the rate of the forward reaction until it equals the reverse reaction.
Industrial Applications
Q.6
Synthesis of ammonia by Haber's process is an exothermic reaction. N2(B) + 312(g) = 2H3(g) AH = -92.46 kJ What should be the possible effect of change of temperature at equilibrium stage?
Correct Verbatim Answer
Effect of temperature change equilibrium stage (i) If temperature is increased • Heat is treated as a product in exothermic reactions. • Increasing temperature shifts equilibrium to the left (reverse direction) to absorb excess heat. This decreases the yield of NH3 so, the product formation is not favoured at high temperature. • Equilibrium constant Ko decreases because forward reaction is less favored. [HI] INH,I AH = 90 kJ mol (i) volume of the container is decreased (iv) chlorine is added (ii) If temperature is decreased • Lower T favors the exothermic direction (forward reaction). • Equilibrium shifts to the right, producing more NH3. • Kc increases because forward reaction is more favored. Optimum conditions (Haber's process) • Too low a temperature gives high yield but very slow rate. • Optimum temperature = 450 °C is used to balance rate and yield.
Q.7
Why ice at 0 °C be melted by applying pressure without supply of heat from outside?
Correct Verbatim Answer
Because ice has a lower density than water. Applying pressure lowers the melting point, causing it to melt even without heat. Write
Q.7
K. for the follow reaction is 0.016 at 520°C 2112(g) H, + 12(g) The equilibrium mixture contains HI = 0.08 M; H2 = 0.01 M and I2 = 0.01M. To this mixture, more HI is added. So that its new concentration is 0.096 M. What will be the concentration of HI, H2 and I2 when equilibrium is re-established?
Correct Verbatim Answer
HI(g) 2(g)+12(g) with Kc = 0.016at520°C Initial equilibrium concentrations (given) [HI] = 00080M, [2] = 0.010M, After adding HI: [HI] becomes 0.096M, while [H,] = 0.010M and [|½] = 0.010M just after addition (before re-equilibration). → Let x be the amount of H2 (and I2) formed as Set up as (after addition, before shift equilibrium shifts to the right): Initial (after addition) Change Equilibrium Write Kc expression: K. - [H, JI21 _ (0.010+x)0.010+x) [HI] Take square root (all concentrations positive), so 0.010+ x - = V0.016 = 0.1265 0.096- 2x 0.010+ x =0.1264911(0.096-2x) 0.010+ x= 0.0121432-0.2529822x x+ 0.2529822x = 0.0121432-0.010 1.2529822x = 0.0021432 x~ 0.001704M Now plug back to get equilibrium concentrations: • [HIle = 0.096 - 2x = 0.096 - 2(0.0017104) = 0.0925791M • [H,]eg = 0.010+ x= 0.0117104M • L2 eq = 0.0117104M [2] = 0.010M. 2HI + 12 H2 0.096 0.010 0.010 +x -2x +x 0.096- 2x 0.010+x 0.010+x - = 0.016 (0.096-2x)2 Final answers: [HIleg = 0.09258M1, 1_ (0.0117104) Check: (0.0925791)= = 0.016 [HI]
Q.8
conditions of equilibrium constant. of
Correct Verbatim Answer
The conditions equilibrium constant are • Temperature must remain constant • It applies only to a system in dynamic equilibrium. The reversible reaction:
Q.8
The equilibrium constant for the reaction between acetic acid and ethyl alcohol is 4. A mixture of 3 moles of acetic acid and 1 mole of ethyl alcohol is allowed to come to equilibrium. Calculate the amount of ethyl acetate present at equilibrium. Reaction
Correct Verbatim Answer
CH, COOH+ C,H2OH CH, COOC, H, + H, O Given Macetic acid initial = 3 mol, Methanolinitial = 1 mol K. = 4, No ester or water initially available, Step 1 - Let x = moles of ester formed at equilibrium At equilibrium Initial moles Species 3 CH, COOH C,H, OH 0 CH, COOCH, 0 H2O Step 2 - Write Kc expression Since this is a liquid-phase reaction (all in same phase), we can use molar amounts directly or concentrations if total volume is constant. The ratio remains same. = [CH, COOC, H, JIH, 0] Substitute equilibrium amounts: 4 = - (3-x) (1 -x) 4 = - (3-x) (1-x) Step 3 - Solve for x Multiply through 4(3-x)(1-x) =x' 4(3-x-3x+x) =x 4(3-4x+x)= x' 12-16x+4x'=x' 12-16x+3x'=0 3x'-16x+12=0 [2]ea=|2lea= 0.01171M Change Equilibrium moles 3 - X - x 1-x +x +x x.X Step 4 - Quadratic equation 16‡/(16) - 4(3)(12) x= 2(3) 161V256-144 x = - 6 16‡ V112 - , x = - 6 Two possibilities 26.583 X = - = 4.43 (impossible - exceeds initial ethanol moles) 6 5.417 = 0.903 • x = 6 x = 0.903 mol Step 5 - Final answer So moles of ethyl acetate at equilibrium = 0.903 mol.
Q.9
2502(g) + 02(g) = 2503(8) has come to equilibrium in a vessel of specific volume at a given temperature. the the reaction Before began, concentrations of the reactants were 0.060 mol/dm? of SO2 and 0.050 mol/dm? of Oz. After equilibrium was reached, the concentration of SO3 was 0.040 mol/dm What is the equilibrium concentration of O2?
Correct Verbatim Answer
We are given the reversible reaction 2502(g) + 02(g) 2503(8) Initial concentrations • [SO'^] = 0.60 mol/dm? • [O,] = 0.50 mol/dm • [SO,] = 0 initially • At equilibrium [SO,] = 0.040 mol/dm SLO BASED SHORT QUESTION ANSWERS
Q.9
Study the equilibrium. 4,0(8) + CO2(g) + CO2(g) (i) Write the expression of Kp. (ii) When 1.00 mole of steam and 1.00 mole of CO are allowed to reach equilibrium, 33.3% of equilibrium mixture is hydrogen. Calculate the value of Kp. State the units of Kp.
Correct Verbatim Answer
Reaction H, (B) + CO (g) =H2(g) +CO2(g) (i) Expression for Kp = K Numerical value of Kp (given data) Initial: 1.00 mol H20 and 1.00 mol CO; no H2 or CO2 initially Let the equilibrium moles of H2 be x. Then at equilibrium: • MH. = X, Nco, = x • "HO = 1-X, Nco =1-x • Total moles = (1-x)+(1-x)+x+x= 2 (constant) Given: 33.3% of the equilibrium mixture is hydrogen, i.e. mole fraction of H2, - = 0.333 Унг = So equilibrium moles Мн. = псо, = 0.666, Mole fractions Ун, = Усо, = 0.666/ 2 = 0.333, 16+10.583 x = - 6 PH, PCoz PH,oPCo x = 0.666 HO = Nco = 0.334 Ун,о = Усо = 0.334/2 = 0.167 Partial pressures P, = y; Potal Substitute into Ap: (Ун, P)(Усо."). (Ун,оP)(Усо") УноУсо (the total pressure P cancels because there are two factors of P in numerator and denominator) Now plug values Ap + (0.333)(0.333) (0.167)(0.167) A quicker algebraic route (same result): since Y, = x/ 2and Yu, = (1-x) / 2, (x/2) A, = ((1-x)/2) With x = 0.666 gives Kp= (0.666/0.334)2 = 3.98 Unit of Kp When An = 0, K, is dimensionless (unitless). K, has no unit. . - Ун,Усо 0.110889 • = - - ~ 3.98(~4.0) 0.027889 -(ix)
Q.10
What is chemical equilibrium?
Correct Verbatim Answer
Chemical equilibrium is a state in a reversible reaction where the rate of forward reaction equals the rate of backward reaction, and the concentrations of reactants and products remain constant. Example = 2NH 3(8) N 2(g) + 3H 2(g) reaches equilibrium when The system ammonia is being formed and decomposed at the same rate.
Q.11
What is a reversible reaction? Give an example.
Correct Verbatim Answer
A reversible reaction is one in which the products can react to form the original reactants. Example = 2H1 (g) 12(g) +12(g) Step 1: Use stoichiometry to find change in concentration. Since the balanced equation shows that 2 moles of SO2 produce 2 moles of SO3, the change in concentration of SO2 is equal to that SO3. Change in [SO,] = - 0.40 mol/dm [SO lea = 0.060-0.040 = 0.020 mol/dm From the balanced equation, 1 mole of O2 is consumed for every 2 moles of SO3 formed, SO: 1 -x0.040 = -0.020 mol/dm? Change in [O2] = 2 [Oleg = 0.050-0:020 = 0.30 mol/dm Answer: [O2]= mol/dm Characteristics of Equilibrium is chemical
Q.12
Why equilibrium called a dynamic equilibrium?
Correct Verbatim Answer
Because at equilibrium, both forward and reverse reactions continue to occur, but at equal rates, resulting in no net change in concentration of reactants and products.
Q.13
Does equilibrium mean equal amounts of reactants and products? Explain.
Correct Verbatim Answer
No equilibrium means equal reaction rates, not equal concentrations. The amounts depend on reaction conditions and the equilibrium constant
Law of Mass Action
Q.14
State the Law of Mass Action.
Correct Verbatim Answer
It states that at constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to the of power their stoichiometric coefficients.
Q.15
How is the equilibrium constant expression written for a reaction?
Correct Verbatim Answer
For the reaction aA + bB == cC + dD The equilibrium constant (Kc) i
Equilibrium Constant
Q.16
What is Kr?
Correct Verbatim Answer
Ko is the equilibrium constant in terms of concentration. It represents the ratio of product concentrations to reactant concentrations at equilibrium.
Q.17
What does the value of Kc indicate about a reaction?
Correct Verbatim Answer
•Kc > 1: Products are favored (reaction goes nearly to completion). • Kc < 1: Reactants are favored (reaction hardly proceeds). • Kc = 1: Significant amounts of both reactants and products present.
Q.18
Is Kc affected by pressure, volume, or concentration?
Correct Verbatim Answer
No. Kc only depends on temperature. or pressure, Changing volume, concentrations affects equilibrium position, not Kc. Types of Equilibria
Q.19
Differentiate between homogeneous and heterogeneous equilibrium.
Correct Verbatim Answer
•Homogeneous equilibrium: All reactants and products are in the same phase. Example: H2(g) + 12(g) == 2H1 (g) • Heterogeneous equilibrium: Reactants and products are in different phases. Example CaCO3(g) = CaO(g) + CO 2(g)
Le Chatelier's Principle
Q.20
State Le Chatelier's Principle.
Correct Verbatim Answer
If a system at equilibrium disturbed by changing temperature, pressure, or concentration, the system shifts its equilibrium position to oppose the change. in an increase does
Q.21
How concentration affect equilibrium?
Correct Verbatim Answer
The system shifts to consume the added species. Example: Adding more H2 in H2(g) + 12(g) shifts equilibrium to the right.
Q.22
What is the effect of pressure on equilibrium in gaseous reactions?
Correct Verbatim Answer
Increasing pressure favours the side with fewer gas molecules; decreasing favors the side with more. Example N2(g) + 3H12(g) = 2N3(g) Increased pressure shifts equilibrium to the right. does temperature affect
Q.23
How equilibrium?
Correct Verbatim Answer
•For exothermic reactions, increasing temperature shifts equilibrium to the left. • For endothermic reactions, increasing temperature shifts equilibrium to the right
Industrial Applications
Q.24
Why is high pressure used in the Haber process? Because the forward reaction
Correct Verbatim Answer
(formation of NH3) involves fewer gas molecules, high pressure favors ammonia production.
Q.25
What is the compromise temperature in the Haber process?
Correct Verbatim Answer
About 450-500°C is used to balance is the rate of reaction and yield because higher temperature decreases yield (due to exothermic nature) but increases rate. Effect of Temperature and Catalyst on Equilibrium
Q.26
What is the role of temperature in exothermic reversible reaction? after
Correct Verbatim Answer
In exothermic reactions, achieving equilibrium, if temperature is increased it favours reverse reaction. If temperature is decreased after attaining equilibrium, it favours forward reaction. In both Ko Will change.
Q.27
What is the effect of change of catalyst on equilibrium?
Correct Verbatim Answer
The change of catalyst will not affect or disturb the equilibrium, as catalyst helps to attain the equilibrium only by equally increasing rate of forward and reverse reaction. in
Q.28
How temperature effects endothermic reversible reaction?
Correct Verbatim Answer
In endothermic reversible reactions, increasing temperature favours forward direction. Decrease in temperature favours reverse direction. In both Kc Will change. Calculations Based on Ko
Q.29
What is the unit of Kc?
Correct Verbatim Answer
Depends on the number of moles. For: aA + bB → cC + dD Unit of Kc=(mol dm3)(c+d) (a+b)
Q.30
What does a small Kc value mean?
Correct Verbatim Answer
The reaction has very low product concentration at equilibrium; the reaction the forward does not proceed much in direction.
Q.31
Why are solids and liquids not included in Ke expression?
Correct Verbatim Answer
Because their concentrations remain constant and do not affect equilibrium.
Q.32
How does a catalyst affect equilibrium?
Correct Verbatim Answer
A catalyst increases the rate of both forward and reverse reactions equally, but does not change the position of equilibrium or the value of Kc. Practice-Based / Examples
Q.33
Write the equilibrium expression for: - 2503(B) 2502(g) + 02(g)
Correct Verbatim Answer
[so,] K. =
Q.34
What will happen if COz is removed from this reaction? CaCO3(8) CaO(s) + CO2(g)
Correct Verbatim Answer
The system will shift to the right to the lost replace tavoring CO2, decomposition of CaCO3.
Q.35
Why is equilibrium important in industrial processes?
Correct Verbatim Answer
Equilibrium controls the yield of products. By adjusting conditions (pressure, temperature), industries maximize product formation (e.g., ammonia, sulfuric acid).
Q.36
How can you increase the yield of an endothermic reaction?
Correct Verbatim Answer
By increasing temperature, because heat acts as a reactant and the equilibrium shifts to the right.
Q.37
Define equilibrium mixture.
Correct Verbatim Answer
A mixture of reactants and products present at equilibrium state.
Q.38
Define equilibrium position. of concentrations
Correct Verbatim Answer
The relative reactants and products at equilibrium, indicating which side is favored
Q.39 Define the reaction quotient (Q):
Correct Verbatim Answer
Q is calculated like Kc, but for non- equilibrium conditions. It helps predict the direction the reaction will proceed to reach equilibrium. DESCRIPTIVE QUESTIONS (EXERCISE)