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PHY./INO. CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 NO. 27 & 28

Course : VIJETA (JP)

This DPP is to be discussed in the week (20.07.2015 to 25.07.2015)

DPP No. # 27 (JEE-MAIN) Total Marks : 60

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.18 Comprehension ('–1' negative marking) Q.19 to Q.20

1.

(3 marks, 2 min.)[54, 36] (3 marks, 2 min.)[06, 04]

The emf of the cell Zn | ZnCl2 (0.05 M) | AgCl(s) | Ag is 1.015 V at 298 K and the temperature coefficient of its emf is – 4.92 × 10–4

V . How many of the reaction thermodynamic parameters G, S and H are negative K

at 298 K? (A) None of them 298 K

(B) One of them

(C) Two of them

(D*) All of them

ij lsy Zn | ZnCl2 (0.05 M) | AgCl(s) | Ag dk emf 1.015 V gS rFkk blds emf dk rkieku xq.kkad

V gSA 298 K ij fdrus Å"ekxfrdh; vfHkfØ;k izkpky G, S rFkk H ds fy, eku _.kkRed gksxsa \ K (A) buesa ls dksbZ ugha (B) buesa ls ,d (C) buesa ls nks (D*) mijksDr lHkh Zn | ZnCl2 (0.05M) | AgCl(s) | Ag Ecell = 1.015V – 4.92 × 10–4

Sol.

 dE cell    = –4.92 × 10–4 V dT   K G = –nFEcell = –ve (as Ecell = +ve) = –2 × 96500 × 1.015 = –195.895 kJ.  dE cell   S = nF   dT  = 2 × 96500 × –4.92 × 10–4 = –94.956 J/K.  G = H – TS. = –198.895 +

298 × (–94.956) = –224.19 KJ. 1000

So, all are negative. blfy, 2.

lHkh _.kkRed gSa

The following compounds have been arranged in order of their increasing thermal stabilities. Identify the correct order. K2CO3 (I) MgCO3 (II) CaCO3 (III) BeCO3 (IV) (A) I < II < III < IV

(B*) IV < II < III < I

(C) IV < II < I < III

(D) II < IV < III < I

fuEufyf[kr ;kSfxdks dks muds rkih; LFkkf;Ro ds c