JEE chemistry

Sponsor Area

Question
CBSEENCH11008038

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by:

  • 2meV

  • square root of mev
  • square root of 2 meV end root
  • mev

Solution

C.

square root of 2 meV end root

The relation between h/λ and energy is given as:
Applying de-Broglie wavelength and kinetic energy term in eV.
de-Broglie wavelength for an electron (λ) = h/p
⇒ p  = h/ λ       (i)
Kinetic energy of an electron = eV
As we know that,
KE space equals space fraction numerator straight p squared over denominator 2 straight m end fraction
therefore space eV space equals space fraction numerator straight p squared over denominator 2 straight m end fraction
space or space straight p space equals square root of 2 meV end root
From equations (i) and (ii), we get
straight h over straight lambda space equals space square root of 2 meV end root

Sponsor Area

Question
CBSEENCH11008039

The heats of combustion of carbon and carbon monoxide are −393.5 and −283.5 kJ mol−1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is:

  • 676.5

  • -676.5

  • -110.5

  • 110 s

Solution

C.

-110.5

C(s) + O2 (g) → CO2 (g); ΔH = -393.5 kJ mol-1 ... (i)
CO + O2/2 → CO2 (g); ΔH = - 283.5 kJ mol-1 ....(ii)
On subtracting Eq. (ii) from Eq. (i), we get
C (s) + O2/2 (g) → CO (g)

ΔH = (-393.5 + 283.5) kJ mol-1 = - -110 kJ mol-1

Question
CBSEENCH11008042

The species in which the N atom is in a state of sp hybridization is:

  • NO2-

  • NO3-

  • NO2

  • NO2+

Solution

D.

NO2+

Question
CBSEENCH11008043

The equilibrium constant at 298 K for a reaction A+B ⇌ C+D is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L−1 ) will be:

  • 0.818

  • 1.818

  • 1.182

  • 0.182

Solution

B.

1.818

space space space space space space space space space space space space space space space space space space space space space straight A space space space space space space plus space space space straight B space rightwards harpoon over leftwards harpoon for space space space space space space space space space of space straight C space space plus space space space space space straight D
subscript Intially space at space straight t space equals 0 space space space space space space space space 1 space space space space space space space space space space space space space space space space 1 space space space space space space space space space space space space space space space space space space space 1 space space space space space space space space space space space space space space space 1 end subscript
to the power of at space equilibrium space space space space space space space 1 minus straight x space space space space space space space space space space 1 minus straight x space space space space space space space space space space space space space space space 1 plus straight x space space space space space space space space space space space 1 plus straight x end exponent
straight K subscript eq space equals space fraction numerator left square bracket CD right square bracket over denominator left square bracket straight A right square bracket left square bracket straight B right square bracket end fraction space equals fraction numerator left parenthesis 1 plus straight x right parenthesis left parenthesis 1 plus straight x right parenthesis over denominator left parenthesis 1 minus straight x right parenthesis left parenthesis 1 minus straight x right parenthesis end fraction space equals space fraction numerator left parenthesis 1 plus straight x right parenthesis squared over denominator left parenthesis 1 minus straight x right parenthesis squared end fraction
Or space 100 space equals space open parentheses fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction close parentheses squared
10 space equals space fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction
or space 10 minus 10 straight x space equals space 1 space plus space straight x
10 minus 1 space space equals space straight x space plus space 10 straight x
9 space equals space 11 space straight x
straight x space equals space 9 over 11 space equals space 0.818 space
left square bracket straight D right square bracket space equals space 1 space plus straight x space equals space 1 space plus space 0.818 space equals space 1.818 space space space

Question
CBSEENCH11008048

Which of the following atoms has the highest first ionisation energy?

  • Na

  • K

  • Sc

  • Rb

Solution

C.

Sc

Order of first ionisation energy is Sc> Na > K > Rb. Due to poor shielding effect, removal of one electron from 4s orbital is difficult as compared to 3s-orbital.