Chemistry

28. (a) Δ_rH^?  is negative, Δ_rS^?  positive, Δ_rG^?  is negative.

27.  (a) The magnitude of the enthalpy change depends on the strength of the intermolecular interactions in the substance undergoing the phase transformations. For example, the strong hydrogen bonds between water molecules hold them tightly in liquid phase. For an organic liquid, such as acetone, the intermolecular dipole-dipole interactions are significantly weaker. Thus, it requires less heat to vaporise 1 mol of acetone than it does to vaporize 1 mol of water.

26.  (b) A system at higher temperature has greater randomness in it than one at lower temperature. Thus, temperature is the measure of average chaotic motion of particles in the system.

25.  (c) When a liquid crystallizes into solid or after freezing, the molecules attain an ordered state and therefore, entropy decreases.

24.  (d) In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and Δ_rH will also be negative. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and Δ_rH will be positive.

23. (a) The state of a thermodynamic system is described by its measurable or macroscopic (bulk) properties. We can describe the state of a gas by quoting its pressure (p), volume (V), temperature (T), amount (n) etc. Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached. Example: Volume of water in a pond, is a state function, because change in volume of its water is independent of the route by which water is filled in the pond, either by rain or by tube well or by both.

22. q_rev = (-Δ_fH^? ) = - (- 286kJ mol^-1) = 286 x 103 Jmol^-1 = 286000 J mol^-1

ΔS  _{(surroundings)} = q_rev/ T = 286000 J mol^-1 / 298 K

                        =  959.73 J mol^-1 K^-1

21. For NO (g), Δ_rH° is a +ve value. So, it is unstable in nature.

For NO₂  (g), Δ_rH° is a -ve value. So, it is stable

20. ΔG° = -RT ln K = - 2.303 RT log K

Putting the values of

R = 8.314 J K^-1 mol^-1,

T = 300 K and

K =10; we get

ΔG° = - 2.303 x 8.314 J K^-1 mol^-1 x 300 K and K x log 10

        = - 5527 J mol^-1

        = -5.527 kJ mol^-1

19. ΔH° = ΔU° + Δ^ng RT

ΔU° = -10.5 kJ, Δ^ng = 2-3 = -1 mol, R = 8.314 x 10^-3 kJ mol^-1, T = 298 K

ΔH° = (- 10.5 kJ) + [ (- 1 mol) x (8.314 x 10^-3 kJ mol^-1) x (298 K)]

       = -10.5 kJ – 2.478 kJ

       = -12.978 kJ

According to Gibbs Helmholtz equation:

ΔG° = ΔH° - TΔS°

        = (- 12.978 kJ) – (298 K) x (- 0.0441 kJ K^-1)

        = -12.978 + 13.142

        = 0.164 kJ

Since the value of ΔG° is positive, the reaction is non-spontaneous.