Equilibrium · Question 3

Given is a concentrated solution of a weak electrolyte AxBy of concentration 'c' and dissociation constant 'K'. The degree of dissociation is given by :

1. [K cx+y-1xxyy ]x+y
2. ( Kcx+y-1xxyy ) 1x+y
3. ( cx+y-1xxyyK )x+y
4. ( cx+y-1xxyyK ) 1x+y

In a closed flask at 600 K, one mole of X2Y4(g) attains equilibrium as given below : X2Y4(g) 2XY2(g) At equilibrium, 75% X2Y4(g) was dissociated and the total pressure is 1 atm. The magnitude of rG (in kJ mol-1) at this temperature is . (Nearest Integer) (Given : R = 8.3 J mol-1 K-1; ln 10 = 2.3, log 2 = 0.3, log 3 = 0.48, log 5 = 0.69, log 7 = 0.84)

M3A2 is a sparingly soluble salt of molar mass y g mol-1 and solubility x g L-1. The ratio of the molar concentration of the anion ( A3-) to the solubility product of the salt is

1. 154 y4x4
2. y5108x4
3. 108 x5y5
4. 1108 y4x4

Arrange the following resultant mixtures in increasing order of their pH values Choose the correct answer from the options given below :

1. B < C < A
2. C < A < B
3. C < B < A
4. A < C < B

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) array|c|c| 1T ( K-1) & 10 Kp \ 0.05 & 3.5 \ 0.06 & 2.5 \ 0.07 & 1.5 \ array The magnitude of H R calculated from the above data is . (Nearest integer)

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 Kp of the reaction can be calculated from the expression :

1. x2a2 + x2 p
2. x2a2 - x2 p
3. a + x2x2 p
4. a2 - x2x2 p

At T(K), the equilibrium constant of A2(g) + B2(g) C(g) is 2.7 10-5. What is the equilibrium constant for 13A2(g) + 13B2(g) 13C(g) at the same temperature?

1. (2.7 10-5)3
2. 6 10-2
3. 2.7 10-5
4. 3 10-2

The pH of a solution obtained by mixing 5 mL of 0.1 M NH4OH solution with 250 mL of 0.1 M NH4Cl solution is 10-2. (Nearest integer) Given: pKb(NH4OH) = 4.74 2 = 0.30 3 = 0.48 5 = 0.70

20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? (pKa value of acetic acid is 4.75).

1. 7.0
2. 4.75
3. 3.5
4. 4.82

For the following reaction at 50 °C and at 2 atm pressure, 2N2O5(g) 2N2O4(g) + O2(g) N2O5 is 50% dissociated. The magnitude of standard free energy change at this temperature is x. x = J mol-1 [Nearest integer]. Given : R = 8.314 J mol-1 K-1, log 2 = 0.30, log 3 = 0.48, ln 10 = 2.303, °C + 273 = K