Thermodynamics

Consider the reaction X Y at 300 K. If H and K are 28.40 kJ mol-1 and 1.8 10-7 at the same temperature, then the magnitude of S for the reaction in J K-1 mol-1 is . (Nearest integer) (Given : R = 8.3 J K-1 mol-1, ln 10 = 2.3, log 3 = 0.48, log 2 = 0.30)

Consider the following data for the reaction X2(g) + Y2(g) 2XY(g) at 600 K. The rG (in kJ mol-1) for the reaction is : array|c|c|c| Compound & fH 600 K ( kJ mol-1) & S 600 K ( J mol-1 K-1) \ XY(g) & 42 & 200 \ X2(g) & 8 & 140 \ Y2(g) & 80 & 250 \ array

1. −21000
2. −10
3. −1000
4. −9.012

At the transition temperature T, A B and G = 105 - 35 T where A and B are two states of substance X. The transition temperature in °C when pressure is 1 atm is . (Nearest integer)

The plot of 10 K vs 1T gives a straight line. The intercept and slope respectively are (where K is the equilibrium constant).

1. - S R2.303, H R2.303
2. - H 2.303R, S 2.303R
3. S 2.303R, - H 2.303R
4. 2.303R H , 2.303R S

Use the following data : array|c|c|c| Substance & f H (500 K) kJ mol-1 & S (500 K) J K-1 mol-1 \ AB(g) & 32 & 222 \ A2(g) & 6 & 146 \ B2(g) & x & 280 \ array One mole each of A2(g) and B2(g) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K. A2(g) + B2(g) 2AB(g) The value of x (in kJ mol-1) is . (Nearest integer) (Given : K = 2.2, R = 8.3 J K-1 mol-1)

Let us consider a reversible reaction at temperature, T. In this reaction, both H and S were observed to have positive values. If the equilibrium temperature is Te, then the reaction becomes spontaneous at:

1. Te > T
2. T > Te
3. T = Te
4. Te = 5T

Which of the following graphs correctly represents the variation of thermodynamic properties of Haber's process?

1. [figure]
2. [figure]
3. [figure]
4. [figure]

Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.

1. Both and are (-ve)
2. is (+ve) but is (-ve)
3. is (-ve) but is (+ve)
4. Both and are (+ve)

Standard entropies of X2, Y2 and XY5 are 70, 50 and 110 J K-1 mol-1 respectively. The temperature in Kelvin at which the reaction 12X2 + 52Y2 XY5 H = -35 kJ mol-1 will be at equilibrium is (Nearest integer)

The standard enthalpy and standard entropy of decomposition of N2O4 to NO2 are 55.0 kJ mol-1 and 175.0 J/K/mol respectively. The standard free energy change for this reaction at 25 in J mol-1 is (Nearest integer)

The effect of temperature on spontaneity of reactions are represented as : [figure] The incorrect combinations are :

1. A and C only
2. B and D only
3. A and D only
4. B and C only

Match List - I with List - II. Choose the correct answer from the options given below :

1. A-I, B-II, C-IV, D-III
2. A-II, B-III, C-I, D-IV
3. A-II, B-I, C-IV, D-III
4. A-II, B-I, C-III, D-IV

When Hvap = 30 kJ/mol and Svap = 75 J mol-1 K-1, then the temperature of vapour, at one atmosphere is K.

For the reaction at 298 K, 2A + B C, H = 400 kJ mol-1 and S = 0.2 kJ mol-1 K-1. The reaction will become spontaneous above K.

For a certain reaction at 300 K, K = 10, then for the same reaction is − 10-1 kJ mol-1. (Given R = 8.314 J K-1 mol-1)

Consider the following reaction at 298 K: 32O2(g) O3(g); KP = 2.47 10-29. rG for the reaction is kJ. (Given R = 8.314 JK-1 mol-1)

Which of the following is not correct?

1. G is positive for a spontaneous reaction
2. G is positive for a non-spontaneous reaction
3. G is zero for a reversible reaction
4. G is negative for a spontaneous reaction

For a certain thermochemical reaction M N at T = 400 K, = 77.2 kJ mol-1, = 122 J K-1, log equilibrium constant (log K) is − 10-1.