Class 11th
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5 months agoContributor-Level 10
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.
New answer posted
5 months agoContributor-Level 10
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.
New answer posted
5 months agoContributor-Level 10
25. (c) When a liquid crystallizes into solid or after freezing, the molecules attain an ordered state and therefore, entropy decreases.
New answer posted
5 months agoContributor-Level 10
24. (d) In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, qp will be negative and ΔrH will also be negative. Similarly in an endothermic reaction, heat is absorbed, qp is positive and ΔrH will be positive.
New answer posted
5 months agoContributor-Level 10
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.
New answer posted
5 months agoContributor-Level 10
22. qrev = (-ΔfH? ) = - (- 286kJ mol-1) = 286 x 103 Jmol-1 = 286000 J mol-1
ΔS (surroundings) = qrev/ T = 286000 J mol-1 / 298 K
= 959.73 J mol-1 K-1
New answer posted
5 months agoContributor-Level 10
21. For NO (g), ΔrH° is a +ve value. So, it is unstable in nature.
For NO2 (g), ΔrH° is a -ve value. So, it is stable
New answer posted
5 months agoContributor-Level 10
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
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