Chemical Equilibrium Chemistry (H) 1St Year SEM-II, Dated: 7Th April 2020

Remarks of Assignment 2: Chemical Equilibrium Chemistry (H) 1st Year SEM-II, Dated: 7th April 2020

Grades: A (Excellent); B+ (Very good); B (Good) and C (Poor) S. NAME Roll No. Grade Remarks No. 1. Sejal Jain 1931210 Resubmission: Not submitted or misplaced 2. Harshita 1931206 A In Q.6. no need to discuss the situations, according to figure 1 it is NO and Q.8. Try to write concisely 3. Yogesh Kumar 1931152 B+ Q. 6. Not correct 4. Pooja 1931132 B+ Q. 6. Not correct 5. Riddhima 120 A Excellent 6. Jyoti 112 B+ Q. 6. Not correct 7. Vivek 146 B+ Q. 6. Not correct 8. Vikash 154 B+ Q. 6. Not correct 9. Nidhi 150 A Excellent 10. Mehak Vaish 136 B+ Q. 6. Not correct 11. Rinka* 122 B+ Q. 6. Not correct 12. Sheenu 202 A Excellent 13. Bhupendra 104 B+ Q. 6. Not correct Maurya 14. Deepanshu 142 B+ Q. 6. Not correct 15. Diksha Rathi 204 A Excellent 16. Khushboo Mittal 256 A Q. 6. Not correct 17. Amit Patwa 252 B+ Q. 6. Not correct 18. Aman Tomar 246 A Excellent 19. Vinay Sharma 176 B+ Q. 6. Not correct 20. Garima Nveen 128 B+ Q. 6. Not correct and Q. 8. Answer in terms of number of moles, 21. Yatin Kumar 254 B+ Q. 6. Not correct 22. Ajay Kumar 238 B+ Q. 6. Not correct 23. Divya Yadav 214 A Excellent, Kindly check Q.6. not properly scanned but I think you want to say NO 24. Kushagra Malik 174 B+ Q. 6. Not correct 25. Megha Dhingra 184 B+ Q. 6. Not correct 26. Prerna Mahajan 186 B+ Q. 6. Not correct 27. Ramkrishna 230 B+ Q. 6. Not correct Bidika 28. Ramsha 218 B+ Q. 6. Not correct 29. Rinku Rathee 236 B+ Q. 6. Not correct 30. Riya Singh 248 B+ Q. 6. Not correct 31. Rohit Pawar 200 A Q. 6. Not correct 32. Saachi Sood 148 B+ Q. 6. Not correct and Q.8. partially correct 33. Sachin Yadav 140 A Q. 6. Not correct 34. Satwik Gupta 130 A Excellent 35. Shivendra Singh 158 B+ Q. 6. Not correct 36. Simran Giri 198 B+ Q. 6. Not correct 37. Somya Chawla 216 B+ Q. 6. Not correct 38. Tarnija Midha 126 A Excellent

39. Yousuf Khan 208 B+ Q. 6. Not correct 40. Mubarak Ali 250 B+ Q. 6. Not correct 41. Vijay Bhasin 172 B+ Q. 6. Not correct 42. Rishav Kumar 118 Resubmission: Not submitted or misplaced 43. Divyansh 226 B+ Q. 6. Not correct Chaudhary 44. Rohit Yadav 234 B+ Q. 6. Not correct 45. Shekhar Yadav 178 B+ Q. 6. Not correct 46. Yesh Vardhan 190 Resubmission: Not submitted or misplaced 47. Abhishek Meena 244 B+ Q. 6. Not correct 48. Abhishek Singh 220 B+ Q. 6. Not correct 49. Ajeet Kumar 242 B+ Q. 6. Not correct 50. Madhusoodan P 224 B+ Q. 6. Not correct Meghwal 51. Chahal Kohli 188 B+ Q. 6. Not correct

Notes: (i) Absent: 182,194, 232, and 240 (Resubmission for all) (ii) Consult Atkins Book for answers.

Answers:

Figure 1. Ans: Q.6. & Q.7.: No, the given figure 1 (pair of weights joined by a string) does not represent a chemical reaction in equilibrium. We can compare the figure 1 with a chemical reaction as:

Explanation: Spontaneity does not represent a chemical reaction in equilibrium. Means, a reaction can be spontaneous but not in equilibrium. Equilibrium means the rate of forward reaction is equal to the rate of backward reaction. As we know that the conditions or criteria for spontaneity and equilibrium are as:

SPONTANEOUS EQUILIBRIUM NON-SPONTANEOUS ∆G < 0 ∆G = 0 ∆G ˃ 0 ∆S ˃ 0 ∆S = 0 ∆S < 0 Q < K Q = K Q ˃ K Where ∆G is free energy of the reaction; ∆S is entropy of the reaction; Q is the reaction quotient and K is the equilibrium constant (See Q.4. of assignment 1 which depicts the relationship between ∆G, Q and K). In figure 1, the heavier weight moves downward (a natural tendency) due to gravity. Means, this downward movement is spontaneous. At the same time, the movement of lighter weight is due to the downward movement of heavier weight and against the gravity. Therefore, the movement of lighter weight is non-spontaneous and its movement is driven by heavier weight. Thus, figure 1 justifies that a spontaneous process drives a non- spontaneous process. Hence, we can say that “Reactions at equilibrium are spontaneous in neither direction: they are neither spontaneous (exergonic) nor non-spontaneous (endergonic)”.

Figure 2.

Ans: Q.8.: The equilibrium constant, K depends on ∆G0, which is defined at standard pressure. The value of ∆G0, and hence of K, is therefore independent of the pressure (P) at which the equilibrium is actually established. Therefore, (∂K/∂P)T = 0, this shows that K is independent of P and this does not necessarily mean that equilibrium composition is independent of P, and its effect depends on how the pressure is applied. Figure 2 justifies this condition.

Mathematically: A ↔ 2 B The equilibrium constant is:

ퟐ ퟎ 푲 = 풑푩 /풑푨풑 ………………… (1)

In this case, since the reaction is at equilibrium, and is compressed from (a) to (b), the system responds by reducing the no. of molecules in gas phase in order to minimize the increase in pressure. Let amount of A present initially = n and no B is present that is,

A ↔ 2 B Initially: n 0 At equilibrium: (1 - α) n 2αn Where α is extent of dissociation of A into 2B. Mole fractions at equilibrium are:

푥 (1−훼)푛 (1−훼) and 퐴= = (1−훼)푛 +2훼푛 (1+훼)

2훼푛 2훼 푥퐵= = (1−훼)푛 +2훼푛 (1+훼) Equation (1) becomes:

ퟐ ퟐ ퟐ ퟐ ퟎ 풑푩 풙푩 푷 ퟒ훼 (푷⁄풑 ) 푲 = ퟎ = ퟎ = ퟐ 풑푨풑 풙푨푷풑 1−훼 where P = total pressure; (Partial pressure = mole fraction × total pressure) 1 1/2 훼 = ( ) 1 + ퟒ푷⁄푲풑ퟎ This formula shows that, even though K is independent of pressure, the amounts of A and B do depend on pressure. It also shows that as 푷 is increased, α decreases, in accord with Le Chatelier’s principle. Note: Most of the answers (except 2-3 students) are without this mathematical derivation. They answered in terms of no. of moles, volumes and pressure which is also right.

Md. Merajul Islam (Name of Teacher) Dated: 21st April 2020