Question

Thermodynamics. 2011 Without using the value of As predict the sign for the following equations and explain why. 1) 2K (S) + F₂ col 2 KF eg) 2) No(1o, is → Nat(aq) +(10 (aq) _3) 2N OCG) + O2(g) → 210, cg) 4 H₂5Cg7 + O2 (g) → So (S) + H₂ Oca Comorks) i) Predit whо bеr 45 iѕ роѕіtіvе оr pa Qofivo £ 0г 0 0 ch of the following processes assuming each occurs Qt constant prossure. Explain why- 1 H₂O(g) → Hoch 21 2 Žn (s) +0,9 2 n o (s) 3) I₂611 - 1219) 4 Ag+ (aq) + Cl-rag) → AGCI(S) 157 N 2 (g) + O₂(g) → 2N Ocg) (omarks)

Hi,can you help answer 2)a) and b).thank you so much for your help

Answer 1

ANSWERS solution j ak(5) + F2 (g) → akt (g) = + (2) Nadlo (5) Nataq) + clog (aq) = + 3) INOCQ) + 02 (5) ► 2NO, (9) = - U Stor-ve 4 H₂S (g) + O₂(g) → S. (s) + H₂O(g) = - J (combustion reactions) Explanation → Randomness in solids are less. → Random ness in liquids are medium Random ness in gases are high. In first reaction the final product is gas so entropy I CAs) increase so the sign will be t e In and in reaction final product is liquid before reaction it was solid. solid to liquid entropy increases so it will be ti In god reaction we calculate the moles. The moles before reaction are greater than the moles after reaction means entropy decreases, so it will be to In 4th reaction the moles before reaction are greater than the moles after reaction. means entropy decreases. So It will be twe.

Q.NO ANSWERS ],H20 (5) — H20 (1) – entropy increase { positive) | ») azn (s) + O2(g) → azno (s) entropy decreases (negetive) 3) 17 (1) > 1,[9) ► entropy increases (positive) (4) Agt (aq) + ch (aq) → Agel (s) entropy decreases (negetive) 5) N, 19) +02 (9) 2NO(g) → entropy decreases (negetive) Bi1 - 2x illo - [ st ,0 + sezo] Shot Selzo - 205, Helo - D5xn Scho - (з усло- ля, ) - ,о Sclzo - (25 Halo - A Soxn) - DS system 2N₂ (g) + 3H2 (g) 2NH₃ (g) Here the reaction occurs spontaneous at this temperature because IG value is negetive. AG=-ve; reaction is spontaneous A Soxn= -197J/k.