+91-8076753736 support@wiredfaculty.com
logo
logo

Equilibrium

Question
CBSEENCH11008387
Wired Faculty App

Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO3 and Conc. H2SO4.In the mixture, nitric acid acts as a/an:

  • reducing agent

  • acid

  • base

  • catalyst

Solution

C.

base

Proton donor is acids and proton acceptor is bases.
Conc. H2SO4 and conc. HNO3 react in the following manner:
HNO3 + H2SO4 → H2NO3+ +HSO4-
H2NO3+ → NO2+ +H2O
Hence, in this reaction HNO3 acts as a base and H2SO4 as an acid.

Question
CBSEENCH11008267
Wired Faculty App

pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product Ksp of Ba(OH)2 is

  • 3.3 x 10-7

  • 5.0 x 10-7

  • 4.0 x 10-6

  • 5.0 x 10-6

Solution

B.

5.0 x 10-7

Given, pH of Ba(OH)2 = 12
pOH = 14-pH
= 14-12 = 2
We know that,
pOH = -log [OH-]
2 =-log [OH-]
[OH-] = antilog (-2)
[OH-] = 1 x 10-2
Ba(OH)2dissolves in water as 
stack Ba left parenthesis OH right parenthesis subscript 2 space left parenthesis straight s right parenthesis with straight s space mol space straight L to the power of negative 1 end exponent below space rightwards harpoon over leftwards harpoon space stack Ba to the power of 2 plus end exponent with straight s below space plus stack 2 OH to the power of minus with 2 straight s below left square bracket OH to the power of minus right square bracket space equals space 2 straight s space equals space 1 space straight x space 10 to the power of negative 2 end exponent left square bracket Ba to the power of 2 plus end exponent right square bracket space equals space fraction numerator left square bracket OH to the power of minus right square bracket over denominator 2 end fraction space equals space fraction numerator 1 space straight x space 10 to the power of negative 2 end exponent over denominator 2 end fraction straight K subscript sp space equals space left square bracket Ba to the power of 2 plus end exponent right square bracket left square bracket OH to the power of minus right square bracket squared space equals space open parentheses fraction numerator 1 space straight x space 10 to the power of negative 2 end exponent over denominator 2 end fraction close parentheses left parenthesis 1 straight x 10 to the power of negative 2 end exponent right parenthesis squared space equals space 0.5 space straight x space 10 to the power of negative 6 end exponent space equals space 5 space straight x space 10 to the power of negative 7 end exponent

Question
CBSEENCH11008225
Wired Faculty App

The addition of a catalyst during a chemical reaction alters which of the following quantities ?

  • Internal energy

  • Enthalpy

  • Activation energy

  • Entropy

Solution

C.

Activation energy

A catalyst is a substance which alters the reaction but itself remains unchanged in the chemical reaction. In a chemical reaction, it provides a new reaction path by the lowering the activation energy barrier.

Question
CBSEENCH11008390
Wired Faculty App

The dissociation constants for acetic acid and HCN at 25o C are 1.5 X 10-5 and 4.5 x 10-10 respectively. The equilibrium constant for the equilibrium,

CN- + CH3COOH  and HCN + CH3COO-

would be

  • 3 x 105

  • 3.0 x 10-5

  • 3.0 x 10-4

  • 3.0 x 104

Solution

D.

3.0 x 104

Given,
CH subscript 3 COOH space space leftwards harpoon over rightwards harpoon space CH subscript 3 COO to the power of minus space space plus space straight H to the power of plus straight K subscript straight a space equals space 1.5 space straight x space 10 to the power of negative 5 end exponent space space space... left parenthesis straight i right parenthesis HCN space rightwards harpoon over leftwards harpoon space straight H to the power of plus space plus space CN to the power of minus semicolon space straight K subscript straight a space equals space 4.5 space straight x space 10 to the power of negative 10 end exponent space.... space left parenthesis iii right parenthesis CN to the power of minus space plus space CH subscript 3 COOH space leftwards harpoon over rightwards harpoon space HCN space plus space CH subscript 3 COO to the power of minus straight K space equals space ? On space Substracting space Eq. space left parenthesis ii right parenthesis space from space Eq space left parenthesis straight i right parenthesis. space we space get CH subscript 3 COOH space plus CN to the power of minus space leftwards harpoon over rightwards harpoon space HCN space plus space CH subscript 3 COO to the power of minus semicolon straight K space equals space straight K subscript straight a over straight K subscript straight a 1 end subscript space equals space fraction numerator 1.5 space straight x space 10 to the power of negative 5 end exponent over denominator 4.5 space straight x 10 to the power of negative 10 end exponent end fraction space equals space 10 to the power of 5 over 3 space equals space 3.33 space straight x space 10 to the power of 4

Sponsor Area

Question
CBSEENCH11008418
Wired Faculty App

The dissociation equilibrium  of gas AB2 can be represented as

2AB2 (g) ⇌ 2AB (g) + B2 (g)
The degree of dissociation is 'x' and is small compared to 1. The expression relating the degree of dissociation  (x) with equilibrium constant Kp and total pressure p is 

  • (2Kp / p)

  • (2Kp/p)1/3

  • (2Kp/p)1/2

  • (Kp/p)

Solution

D.

(Kp/p)

Initial              1                   0               0
at equ        2(1-x)              2x               x
where,  x = degree of dissociation
Total moles at equilibrium = 2-2x + 2x + x
 = ( 2 + x)
straight p subscript AB subscript 2 end subscript space equals space fraction numerator 2 space left parenthesis 1 minus straight x right parenthesis straight p over denominator left parenthesis 2 plus straight x right parenthesis end fraction straight p subscript AB space equals space fraction numerator 2 xp over denominator left parenthesis space 2 space plus straight x right parenthesis end fraction straight p subscript straight B subscript 2 space end subscript space equals space fraction numerator xp over denominator space left parenthesis space 2 space plus straight x right parenthesis end fraction straight K subscript straight p space equals space fraction numerator left parenthesis straight p subscript AB right parenthesis squared left parenthesis straight p subscript straight B subscript 2 right parenthesis over denominator left parenthesis straight p subscript AB subscript 2 end subscript right parenthesis end fraction space equals space fraction numerator open parentheses begin display style fraction numerator 2 xp over denominator 2 plus straight x end fraction end style close parentheses squared open parentheses begin display style fraction numerator straight x over denominator 2 space plus straight x end fraction end style straight p close parentheses over denominator open parentheses begin display style fraction numerator 2 left parenthesis 1 minus straight x right parenthesis over denominator 2 plus straight x end fraction end style straight p close parentheses end fraction equals space fraction numerator straight x cubed straight p over denominator left parenthesis 2 plus straight x right parenthesis left parenthesis 1 minus straight x right parenthesis squared end fraction open square brackets therefore space straight x less than less than less than space 1 space and space 2 space so comma space left parenthesis 1 minus straight x right parenthesis space almost equal to 1 comma space left parenthesis 2 plus straight x right parenthesis space almost equal to space 2 close square brackets equals fraction numerator straight x cubed straight p over denominator 2 end fraction straight x space equals open parentheses fraction numerator 2 straight K subscript straight p over denominator straight p end fraction close parentheses to the power of 1 divided by 3 end exponent

Sponsor Area

Sponsor Area

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation