JEE Chemistry

Question 1

The incorrect expression among the following is

  • fraction numerator increment space straight G subscript system over denominator increment straight G subscript total end fraction space equals space minus straight T
  • In isothermal process

    straight W subscript reversible space equals negative nRT space ln straight V subscript straight f over straight V subscript straight i

  • In space straight K space equals space fraction numerator increment straight H to the power of straight o minus straight T increment straight S over denominator RT end fraction
  • straight K space equals space straight e to the power of increment straight G to the power of straight o divided by RT end exponent

Solution

C.

In space straight K space equals space fraction numerator increment straight H to the power of straight o minus straight T increment straight S over denominator RT end fraction

Option C has incorrect expression. The correct expression is,
increment straight G to the power of straight o space equals space increment straight H to the power of straight o space minus straight T increment straight S to the power of straight o
increment straight G to the power of straight o space equals nRT space log space straight K
therefore equals negative RT space log space straight K space equals increment straight H to the power of straight o minus straight T increment straight S to the power of straight o
therefore space log space straight K space equals space minus space open parentheses fraction numerator increment straight H to the power of straight o minus straight T increment straight S to the power of straight o over denominator RT end fraction close parentheses

Question 2

The species which can best serve as an initiator for the cationic polymerization is

  • LiAlH4

  • HNO3

  • AlCl3

  • BaLi

Solution

C.

AlCl3

Electron-deficient species (Lewis acid) like AlCl3, BF3, etc, are is used as an initiator for cationic polymerisation.

Question 3

The standard reduction potentials for Zn2+/ Zn, Ni2+/ Ni, and F2+/ Fe are –0.76, –0.23 and –0.44 V respectively. The reaction X + Y2+ → X 2+ + Y will be spontaneous when

  • X = Ni, Y = Fe

  • X = Ni, Y = Zn

  • X =Fe, Y= Zn

  • X = Zn, Y = Ni

Solution

D.

X = Zn, Y = Ni

X = Zn, Y = Ni
Zn + Ni2+  →Zn2+ + Ni
straight E subscript cell superscript straight o space equals space straight E subscript zn divided by zn to the power of 2 plus end exponent end subscript superscript straight o space plus space straight E subscript Ni to the power of 2 plus end exponent divided by Ni end subscript superscript straight o
space equals space 0.76 minus 0.23
space equals space 0.53 space straight V
straight E subscript cell superscript 0 space greater than thin space 0

Question 4

The equilibrium constant (Kc) for the reaction, N2(g) + O2 (g) → 2NO (g) at temperature T is 4 x 10-4. The value of Kc for the reaction NO space left parenthesis straight g right parenthesis space rightwards arrow 1 half space straight N subscript 2 space left parenthesis straight g right parenthesis space plus 1 half straight O subscript 2 space left parenthesis straight g right parenthesis at the same temperature is

  • 0.02

  • 2.5 x 102

  • 4 x 10-4

  • 50.0

Solution

D.

50.0

straight N subscript 2 space left parenthesis straight g right parenthesis space plus space straight O subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow space 2 NO space left parenthesis straight g right parenthesis
straight K subscript straight c space equals space fraction numerator left square bracket NO right square bracket squared over denominator left square bracket straight N subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket end fraction space equals space 4 space straight x 10 to the power of negative 4 end exponent
2 space NO left parenthesis straight g right parenthesis space rightwards arrow space straight N subscript 2 space left parenthesis straight g right parenthesis space plus space straight O subscript 2 space left parenthesis straight g right parenthesis
straight K subscript straight c apostrophe space equals space 1 over straight K subscript straight c space equals space fraction numerator left square bracket straight N subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket over denominator left square bracket NO subscript 2 right square bracket end fraction space equals space fraction numerator 1 over denominator space 4 space straight x 10 to the power of negative 4 end exponent end fraction space equals space 10 to the power of 4 over 4
NO space left parenthesis straight g right parenthesis space rightwards arrow 1 half space straight N subscript 2 space left parenthesis straight g right parenthesis space plus space 1 half space straight O subscript 2 space left parenthesis straight g right parenthesis
straight K subscript straight c apostrophe apostrophe space equals space fraction numerator left square bracket straight N subscript 2 right square bracket to the power of 1 divided by 2 end exponent left square bracket straight O subscript 2 right square bracket over denominator left square bracket NO subscript 2 right square bracket squared end fraction
space equals space square root of straight K subscript straight c apostrophe end root space equals space square root of 10 to the power of 4 over 4 end root
space equals space 100 over 2 space equals space 50
Question 5

The compressibility factor for a real gas at high pressure is

  • 1 plus RT over pb
  • 1
  • 1 plus pb over RT
  • 1 minus pb over RT

Solution

C.

1 plus pb over RT

Vander Waal's equation for one mole of real gas is
open parentheses straight p plus straight a over straight V squared close parentheses space left parenthesis straight v minus straight b right parenthesis space equals space RT
When space pressure space is space high space straight p space greater than greater than straight a over straight V squared
such space that space open parentheses straight p space plus straight a over straight V squared close parentheses space equals space straight p
thus comma space straight p left parenthesis straight v minus straight b right parenthesis space equals space RT
pV minus pb space equals space RT
pv space equals space RT space plus pb
therefore space COmpressibillity space factor comma
straight z space equals space pV over RT space equals space open parentheses 1 plus pb over RT close parentheses