Chapter 08: Chemical Equilibrium

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.

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

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

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.

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.

Because the concentration of reactants decreases, lowering the rate of the forward reaction until it equals the reverse reaction.

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.

Because ice has a lower density than water. Applying pressure lowers the melting point, causing it to melt even without heat. Write

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]

The conditions equilibrium constant are • Temperature must remain constant • It applies only to a system in dynamic equilibrium. The reversible reaction:

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.

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

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)

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.

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

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.

No equilibrium means equal reaction rates, not equal concentrations. The amounts depend on reaction conditions and the equilibrium constant

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.

Ko is the equilibrium constant in terms of concentration. It represents the ratio of product concentrations to reactant concentrations at equilibrium.

No. Kc only depends on temperature. or pressure, Changing volume, concentrations affects equilibrium position, not Kc. Types of Equilibria

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

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

(formation of NH3) involves fewer gas molecules, high pressure favors ammonia production.

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

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.

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

In endothermic reversible reactions, increasing temperature favours forward direction. Decrease in temperature favours reverse direction. In both Kc Will change. Calculations Based on Ko

The reaction has very low product concentration at equilibrium; the reaction the forward does not proceed much in direction.

Because their concentrations remain constant and do not affect equilibrium.

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

Equilibrium controls the yield of products. By adjusting conditions (pressure, temperature), industries maximize product formation (e.g., ammonia, sulfuric acid).

By increasing temperature, because heat acts as a reactant and the equilibrium shifts to the right.

A mixture of reactants and products present at equilibrium state.

The relative reactants and products at equilibrium, indicating which side is favored

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)