Question 1

Consider the following reactions in which all the reactants and products are present in gaseous state 2xy x_2 + y_2 K_1 = 2.5 10^5 xy + 1 2 z_2 xyz K_2 = 5 10^ -3 The value of K_3 for the equilibrium

Accepted Answer

The given reactions are: 2xy x_2 + y_2 K_1 = 2.5 10^5 xy + 1 2 z_2 xyz K_2 = 5 10^ -3 We need to find the equilibrium constant K_3 for the reaction: 1 2 x_2 + 1 2 y_2 + 1 2 z_2 xyz Reversing the first reaction and multiplying it by 1 2 , we get: 1 2 x_2 + 1 2 y_2 xy The equilibrium constant for this modified reaction is: K' = ( 1 K_1 )^ 1 2 = 1 K_1 Adding this modified reaction to the second reaction: 1 2 x_2 + 1 2 y_2 xy (K') xy + 1 2 z_2 xyz (K_2) ----------------------------------------------

Question 2

Solid carbon, CaO and CaCO_3 are mixed and allowed to attain equilibrium at T K. CaCO _3(s) CaO (s) + CO _2(g) K_ p_1 = 0.08 atm C (s) + CO _2(g) 2 CO (g) K_ p_2 = 2 atm The partial pressure of CO is

Accepted Answer

For the first equilibrium: CaCO _3(s) CaO (s) + CO _2(g) K_ p_1 = P_ CO _2 = 0.08 atm For the second equilibrium: C (s) + CO _2(g) 2 CO (g) K_ p_2 = (P_ CO )^2 P_ CO _2 = 2 atm Substituting the value of P_ CO _2 : (P_ CO )^2 0.08 = 2 (P_ CO )^2 = 0.16 P_ CO = 0.4 atm = 4 10^ -1 atm Answer: 4

Question 3

In a closed flask at 600 K, one mole of X_2Y_4(g) attains equilibrium as given below : X _2 Y _4(g) 2 XY _2(g) At equilibrium, 75\% X_2Y_4(g) was dissociated and the total pressure is 1 atm. The magni

Accepted Answer

The given equilibrium reaction is: X _2 Y _4(g) 2 XY _2(g) Initial moles: 1 Degree of dissociation, = 0.75 Moles of X _2 Y _4 at equilibrium = 1 - = 1 - 0.75 = 0.25 Moles of XY _2 at equilibrium = 2 = 2 0.75 = 1.5 Total moles at equilibrium = 0.25 + 1.5 = 1.75 Given total pressure, P = 1 atm. Partial pressure of X _2 Y _4, P_ X _2 Y _4 = 0.25 1.75 1 = 1 7 atm Partial pressure of XY _2, P_ XY _2 = 1.5 1.75 1 = 6 7 atm The equilibrium constant K_p is given by: K_p = (P_ XY _2 )^2 P_ X _2 Y _4 = (

Question 4

One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium. A(g) B(g). K_c for this reaction at 400 K is 4.0. The partial pressures (in at

Accepted Answer

The given reaction is A(g) B(g). Let the initial moles of A be 1 and at equilibrium, let x moles of A dissociate. Moles at equilibrium: n_A = 1 - x n_B = x Since n_g = 0, the equilibrium constant K_c can be written in terms of moles: K_c = n_B n_A = x 1 - x = 4.0 x = 4 - 4x 5x = 4 x = 0.8 At equilibrium, the moles of the gases are: n_ He = 1 n_B = 0.8 Using the ideal gas equation P = nRT V , the partial pressures are calculated as follows: Partial pressure of He: P_ He = 1 0.082 400 10 = 3.28 at

Question 5

The reaction A(g) B(g) + C(g) was initiated with the amount 'a' of A(g). At equilibrium it is found that the amount of A(g) remaining is (a - x) at a total pressure of p. The equilibrium constant K_p

Accepted Answer

The reaction is A(g) B(g) + C(g) Initial moles: a for A, 0 for B, 0 for C Moles at equilibrium: (a - x) for A, x for B, x for C Total moles at equilibrium = (a - x) + x + x = a + x Partial pressures at equilibrium: p_A = a - x a + x p p_B = x a + x p p_C = x a + x p The equilibrium constant K_p is given by: K_p = p_B p_C p_A K_p = ( x a + x p ) ( x a + x p ) a - x a + x p K_p = x^2 (a + x)(a - x) p K_p = x^2 a^2 - x^2 p Answer: x^2 a^2 - x^2 p

Question 6

The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below A(g) B(g) + C(g) 1 T ( K ^ -1 ) _ 10 K_p 0.05 3.5 0.06 2.5 0

Accepted Answer

According to the van 't Hoff equation: _ 10 K_p = - H^ 2.303 R ( 1 T ) + S^ 2.303 R This represents a linear equation y = mx + c, where y = _ 10 K_p and x = 1 T . The slope m of the line is given by: m = - H^ 2.303 R From the given data table, we can calculate the slope m using any two points, for example (0.05, 3.5) and (0.06, 2.5): m = y_2 - y_1 x_2 - x_1 = 2.5 - 3.5 0.06 - 0.05 = -1.0 0.01 = -100 Equating the calculated slope to the theoretical slope: - H^ 2.303 R = -100 H^ R = 100 2.303 = 23

Question 7

For the following reaction at 50°C and at 2 atm pressure, 2N_2O_5(g) 2N_2O_4(g)+O_2(g) N_2O_5 is 50\% dissociated. The magnitude of standard free energy change at this temperature is x. x= ______ J mo

Accepted Answer

Let the initial moles of N_2O_5 be 2. The given reaction is: 2N_2O_5(g) 2N_2O_4(g) + O_2(g) Since N_2O_5 is 50\% dissociated (degree of dissociation = 0.5), the moles at equilibrium are: Moles of N_2O_5 = 2(1 - 0.5) = 1 Moles of N_2O_4 = 2(0.5) = 1 Moles of O_2 = 0.5 Total moles at equilibrium = 1 + 1 + 0.5 = 2.5 Given the total pressure P = 2 atm, the partial pressures of the gases are: P_ N_2O_5 = 1 2.5 2 = 0.8 atm P_ N_2O_4 = 1 2.5 2 = 0.8 atm P_ O_2 = 0.5 2.5 2 = 0.4 atm The equilibrium cons

Question 8

At T(K), the equilibrium constant of A_2(g) + B_2(g) C(g) is 2.7 10^ -5 . What is the equilibrium constant for 1 3 A_2(g) + 1 3 B_2(g) 1 3 C(g) at the same temperature?

Accepted Answer

For the reaction A_2(g) + B_2(g) C(g), the equilibrium constant is K_1 = 2.7 10^ -5 . When a reaction is multiplied by a factor n, the new equilibrium constant becomes K^n. The given reaction is multiplied by 1 3 to obtain the reaction 1 3 A_2(g) + 1 3 B_2(g) 1 3 C(g). The new equilibrium constant K_2 is given by: K_2 = (K_1)^ 1/3 K_2 = (2.7 10^ -5 )^ 1/3 K_2 = (27 10^ -6 )^ 1/3 K_2 = 3 10^ -2 Answer: 3 10^ -2

Question 9

Observe the following equilibrium in a 1 L flask. A ( ~g ) B ( ~g ) At T ( K ), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and

Accepted Answer

Initial equilibrium: [A]=0.5 M, [B]=0.375 M; K_c= 0.375 0.5 =0.75 After adding 0.1 mol A: [A]=0.6 M, [B]=0.375 M; Q=0.625 Q Let x be A that decomposes: 0.375+x 0.6-x =0.75 0.375+x=0.45-0.75x 1.75x=0.075; x≈0.0429 New [A]=0.6-0.0429=0.557 M; [B]=0.375+0.0429=0.418 M

Question 10

Consider the following gaseous equilibrium in a closed container of volume ' V ' at T ( K ). P _ 2 ( ~g )+ Q _ 2 ( ~g ) 2 PQ ( g ) 2 moles each of P _ 2 ( ~g ), Q _ 2 ( ~g ) and PQ ( g ) are present a

Accepted Answer

Initial equilibrium: 2 moles each of P_2, Q_2, PQ with K = [PQ]^2 [P_2][Q_2] = 4 4 = 1. After adding 1 mole each of P_2 and Q_2: initial composition is (3, 3, 2). Reaction quotient Q = 4 9 Let x moles of reactants combine: (2+2x)^2 (3-x)^2 = 1 2+2x = 3-x 3x = 1 → x = 1 3 Final moles: P_2 = 3 - 1 3 = 2.67, Q_2 = 2.67, PQ = 2 + 2 3 = 2.67

Question 11

X _ 2 ( ~g )+ Y _ 2 ( ~g ) 2 Z ( g ) X _ 2 ( ~g ) and Y _ 2 ( ~g ) are added to a 1 L flask and it is found that the system attains the above equilibrium at T ( K ) with the number of moles of X _ 2 (

Accepted Answer

The equilibrium constant for the reaction X_2(g) + Y_2(g) 2Z(g) is calculated from the initial equilibrium state: K_c = [Z]^2 [X_2][Y_2] = 9^2 3 3 = 9 mol/L. When 10 mol of Z(g) is added, the new concentrations are: [X_2] = 3, [Y_2] = 3, [Z] = 19 mol. The reaction quotient is Q = 19^2 3 3 = 40.11. Since Q > K_c, the equilibrium shifts left (toward reactants). Let x mol of Z decompose. At the new equilibrium: [X_2] = 3 + x 2 , [Y_2] = 3 + x 2 , [Z] = 19 - x. Applying the equilibrium expression: (

Question 12

Consider the general reaction given below at 400 K x ~A ( ~g ) y ~B ( ~g ) The values of K _ p and K _ c are studied under the same condition of temperature but variation in x and y. (i) K _ p =85.87

Accepted Answer

The relationship between K_p and K_c is given by K_p = K_c(RT)^ n_g , where n_g = y - x. Given T = 400 K and R = 0.0821 L atm K^ -1 mol^ -1 . The value of RT = 0.0821 400 = 32.84. For case (i): K_p = 85.87 and K_c = 2.586. K_p K_c = 85.87 2.586 33.2. Since 33.2 32.84^1, we have n_g = y - x = 1. Checking options for (i): (1) y-x = 3-1 = 2 (2) y-x = 1-4 = -3 (3) y-x = 1-3 = -2 (4) y-x = 2-1 = 1. This matches. For case (ii): K_p = 0.862 and K_c = 28.62. K_p K_c = 0.862 28.62 0.0301. Since 0.0301 1

Question 13

For the following gas phase equilibrium reaction at constant temperature, NH _ 3 ( ~g ) 1 / 2 ~N _ 2 ( ~g )+3 / 2 H _ 2 ( ~g ) if the total pressure is 3 ~atm and the pressure equilibrium constant ( K

Accepted Answer

For the reaction: NH_3(g) 1 2 N_2(g) + 3 2 H_2(g) At t = 0: 1 mole, –, – At equilibrium: 1- , 2 , 3 2 Total moles = 1 + K_P = ( 2 )^ 1/2 ( 3 2 )^ 3/2 (1- ) ( P_T 1+ )^1 Given P_T = 3 atm and K_P = 9 atm: 9 = ( 2 )^ 1/2 ( 3 2 )^ 3/2 (1- ) (3)^ 1/2 1+ Simplifying: 9 = 9 ( 2 )^2 1- ^2 1 - ^2 = ^2 4 5 ^2 4 = 1 ^2 = 0.8 = (0.8)^ 1/2 = ( 1 0.8 )^ -1/2 = (125 10^ -2 )^ -1/2 So x = 125.

Question 14

Dissociation of a gas A _ 2 takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K. A _ 2 ( ~g ) 2 ~A ( ~g ) The standard Gibbs energy of format

Accepted Answer

For the dissociation reaction A_2(g) 2A(g), first calculate the standard Gibbs free energy: G°_ rxn = 2(-50.832) - (-100.00) = -1.664 kJ/mol. Using G° = -RT K_p: -1664 = -8 300 K_p, giving K_p = e^ 0.6933 = 2. For dissociation with degree of dissociation , starting with 1 mole of A_2 at 1 bar total pressure: At equilibrium, moles are (1- ) for A_2 and 2 for A, with total (1+ ) moles. K_p = 4 ^2 1- ^2 = 2. Solving: 2(1- ^2) = 4 ^2 gives ^2 = 1/3 or = 0.577. Since degree of dissociation is (x 10^

Question 15

For the reaction, N _ 2 O _ 4 2 NO _ 2 , graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is -5.40 ~kJ ~mol ^ -1 . B. As G ^ in graph is po

Accepted Answer

From the graph of G vs fraction of N_2O_4 dissociated: At x = 0 (pure N_2O_4): G = 0 At equilibrium (x 0.4): G = -0.84 kJ/mol (minimum) At x = 1 (complete dissociation): G = 5.40 kJ/mol Statement A: G° = 5.40 kJ/mol (positive), not -5.40 kJ/mol. False. Statement B: Even with positive G°, equilibrium is established with partial dissociation. False. Statement C: Reverse reaction reaches equilibrium, not completion. False. Statement D: G = G_ eq - G_ N_2O_4 = -0.84 - 0 = -0.84 kJ/mol. True. Stateme

Chemical Equilibrium — Previous Year Questions & Solutions