NEET chemistry

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Question
CBSEENCH11008288

Give that the equilibrium constant for the reaction,

2SO2 (g) + O2 (g)  ⇌ 2SO3 (g)

has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction t the same temperature?

SO3 (g)  ⇌ SO2 (g) +1/2 O2 (g)

  • 1.8 x 10-3

  • 3.6 x 10-3

  • 6.0 x 10-2

  • 1.3 x 10-5

Solution

C.

6.0 x 10-2

2SO2 (g) +O2 (g)  ⇌ 2SO3 (g)
Equilibrium constant for this reaction,
straight K equals space fraction numerator left square bracket SO subscript 3 right square bracket squared over denominator left square bracket SO subscript 2 right square bracket squared left square bracket straight O subscript 2 right square bracket end fraction
SO subscript 3 space left parenthesis straight g right parenthesis space space rightwards harpoon over leftwards harpoon space SO subscript 2 space left parenthesis straight g right parenthesis space plus 1 half space straight O subscript 2 space left parenthesis straight g right parenthesis
Equilibrium space constant space for space this space reaction
straight K apostrophe space equals space fraction numerator left square bracket SO subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket to the power of begin display style 1 half end style end exponent over denominator left square bracket SO subscript 3 right square bracket end fraction
On space squaring space eq. space left parenthesis ii right parenthesis thin space both space sides comma space we space have space
left parenthesis straight K to the power of apostrophe right parenthesis squared space equals space fraction numerator left square bracket SO subscript 2 right square bracket left square bracket straight O subscript 2 right square bracket over denominator left square bracket SO subscript 3 right square bracket end fraction
space equals space 1 over straight K
equals 1 over 278
straight K apostrophe space equals space square root of 1 over 278 end root
equals square root of 0.003597 end root
equals space 5.99 space straight x space 10 to the power of negative 2 end exponent
almost equal to space 6 space straight x space 10 to the power of negative 2 end exponent

Sponsor Area

Question
CBSEENCH12011024

Given the reaction between two gases represented by A2 and B2 to give the compound AB (g).
A2(g) +B2 (g)  ⇌ 2AB (g)
At equilibrium the concentration
of A2 = 3.0 x 10-3 M
of B2 = 4.2 x 10-3 M
of AB = 2.8 x 10-3 M

If the reaction takes place in a sealed vessel at 527oC, then the value of Kc will be

  • 2.0

  • 1.9

  • 0.62

  • 4.5

Solution

C.

0.62

A2 (g) +B2 (g)  ⇌ 2AB (g)
The equilibrium constant is given by
straight K subscript straight c space equals space fraction numerator left square bracket AB right square bracket squared over denominator left square bracket straight A subscript 2 right square bracket left square bracket straight B subscript 2 right square bracket end fraction space equals
straight K subscript straight c space equals space fraction numerator left parenthesis 2.8 space straight x space 10 to the power of negative 3 end exponent right parenthesis squared over denominator left parenthesis 3.0 space straight x space space 10 to the power of negative 3 end exponent right parenthesis left parenthesis 4.2 space straight x space 10 to the power of negative 3 end exponent right parenthesis end fraction
straight K subscript straight c space equals space fraction numerator 7.84 over denominator 12.6 end fraction
space equals space 0.62

Question
CBSEENCH12011025

During the change of O2 to O2- ion, the electron adds on which one of the following orbitals?

  • π* orbital

  • π orbital

  • σ* orbital

  • σ orbital

Solution

A.

π* orbital

During the change of O2 to O2-the electron adds on π* orbital. Molecular orbital configuration of O2 (total e- =16)
  straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma space straight sigma 2 straight s squared comma straight sigma asterisk times 2 straight s squared comma straight sigma 2 straight p subscript straight z superscript 2 comma space straight pi 2 straight p subscript straight x superscript 2 almost equal to straight pi 2 straight p subscript straight y superscript 2 space comma space straight pi asterisk times 2 straight p subscript straight x superscript 1 almost equal to straight pi asterisk times 2 straight p subscript straight y superscript 1

and space configuration space of space straight O subscript 2 superscript minus space left parenthesis 8 plus 8 plus 1 space equals space 17 right parenthesis
straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma space straight sigma 2 straight s squared comma straight sigma asterisk times 2 straight s squared comma straight sigma 2 straight p subscript straight z superscript 2 comma space straight pi 2 straight p subscript straight x superscript 2 almost equal to straight pi 2 straight p subscript straight y superscript 2 space comma space straight pi asterisk times 2 straight p subscript straight x superscript 2 almost equal to straight pi asterisk times 2 straight p subscript straight y superscript 1

Question
CBSEENCH11008289

A  certain gas takes three times as long to effuse out as helium. Its molecular mass will be 

  • 27 u

  • 36 u 

  • 64 u

  • 9 u

Solution

B.

36 u 

From Graham's diffusion law,
straight r subscript 1 over straight r subscript 2 space equals space square root of M subscript 2 over M subscript 1 end root
O r
V subscript 1 over t subscript 1 space x space t subscript 2 over V subscript 2 space equals space square root of M subscript 2 over M subscript 1 end root
because space V o l u m e space i s space s a m e
therefore space 3 over 1 equals space square root of straight M subscript 2 over 4 end root
space or space 9 equals straight M subscript 2 over 4
Or space straight M subscript 2 space equals space 9 space straight x space 4 space equals space 36 space straight u

Question
CBSEENCH11008290

The orbital angular momentum of p-electron is given as

  • fraction numerator straight h over denominator square root of 2 straight pi end root end fraction
  • square root of 3 fraction numerator h over denominator 2 pi end fraction
  • square root of 3 over 2 end root space h over pi
  • square root of 6. end root fraction numerator h over denominator 2 pi end fraction

Solution

A.

fraction numerator straight h over denominator square root of 2 straight pi end root end fraction

Orbital angular momentum = 
square root of straight l left parenthesis straight l plus 1 right parenthesis end root space x fraction numerator h over denominator 2 pi end fraction
therefore space F o r space p minus e l e c t r o n comma space l italic space italic equals italic 1
italic therefore italic space O r b i t a l italic space a n g u l a r italic space m o m e n t u m italic comma
italic equals italic space italic space square root of l italic left parenthesis l italic plus italic 1 italic right parenthesis end root space x fraction numerator h over denominator italic 2 pi end fraction
italic equals italic space square root of italic 2 italic space x fraction numerator h over denominator italic 2 pi end fraction
italic equals fraction numerator h over denominator square root of italic 2 pi end fraction