JEE chemistry

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Question
CBSEENCH11008096

The standard enthalpy of formation of NH3 is– 46.0 kJmol–1. If the enthalpy of formation of H2 from its atoms is – 436 kJ mol–1 and that of N2 is – 712 kJ mol–1,the average bond enthalpy of N – H bond is NH3 is

  • -964 kJ mol-1

  • +352 kJ mol-1

  • +1056 kJ mol-1

  • -1102 kJ mol-1

Solution

B.

+352 kJ mol-1

NH subscript 3 space left parenthesis straight g right parenthesis space rightwards arrow with space on top space 1 half space straight N subscript 2 space left parenthesis straight g right parenthesis space space plus space 3 over 2 space straight H subscript 2 space left parenthesis straight g right parenthesis
increment straight H degree space equals space minus space increment minus straight H degree subscript straight f space left parenthesis NH subscript 3 right parenthesis space equals space minus space left parenthesis negative 46 right parenthesis space equals space 46 space kJ space mol to the power of negative 1 end exponent
Also comma space increment straight H degree space equals space 3 increment straight H subscript straight N minus straight H end subscript space plus 1 half increment straight H subscript straight N identical to straight N end subscript space plus 3 over 2 space increment straight H subscript straight H minus straight H end subscript
space 46 space equals space 3 space increment straight H subscript straight N minus straight H end subscript space plus 1 half space left parenthesis negative 712 right parenthesis space plus space 3 over 2 space left parenthesis negative 436 right parenthesis
increment straight H subscript straight N minus straight H end subscript space equals space 1 third left square bracket 1056 right square bracket space equals space plus 352 space kJ space mol to the power of negative 1 end exponent

Sponsor Area

Question
CBSEENCH11008097

Consider the reaction:
Cl2(aq) + H2S(aq) → S (s) + 2H+ (aq) + 2Cl- (aq)
The rate equation for this reaction is
I. Cl2 + H2S → H+ +Cl- + Cl+ +HS-
II. H2S ⇌ H+ + HS- (fast equilibrium)
Cl2 + HS- → 2Cl- + H+ + S (slow)

  • II only

  • Both (I) and (II)

  • Neither (I) nor (II)

  • (I) only

Solution

D.

(I) only

For (A)
rate = K[Cl2] [H2S]
For (B)
rate = K[Cl2] [HS-] … (i)
putting space in space equation space left parenthesis straight i right parenthesis
rate space equals space straight K left square bracket Cl subscript 2 right square bracket space Keq space fraction numerator left square bracket straight H subscript 2 straight S right square bracket over denominator left square bracket straight H to the power of plus right square bracket end fraction

Question
CBSEENCH11008098

If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established?(Given: Vapour pressure of H2O at 300 K is 3170 Pa; R = 8.314 J K–1 mol–1)

  • 5.56 x 10-3 mol

  • 1.53 x 10-2 mol

  • 4.46 x 10-2 mol

  • 1.27 x 10-3 mol

Solution

D.

1.27 x 10-3 mol

The volume occupied by water molecules in vapour phase is (1-10-4) dm3, that is approximately (1 x 10-3)m3
pvapV = nH2O (mol)
3170 (pa) x 1 x 10-3 (m3) - nH2O(mol) x 8.314 (JK-1 mol-1) x 300K

Question
CBSEENCH11008099

The energy required to break one mole of Cl— Cl bonds in Cl2 is 242 kJ mol. The longest wavelength of light capable of breaking a single Cl — Cl bond is
(c= 3 x 108 ms–1and NA = 6.02 x 1023 mol–1)

  • 594 nm

  • 640 nm

  • 700 nm

  • 494

Solution

D.

494

Energy, E = NA
hv space equals space straight N subscript straight A space straight x space hc over straight lambda
or space straight lambda space equals space straight N subscript straight A space straight x space hc over straight E
straight lambda space equals space fraction numerator 6.626 space straight x space 10 to the power of negative 34 end exponent space straight x space 3 space straight x space 10 to the power of 8 space straight x space 6.02 space space straight x space 10 to the power of 23 over denominator 242 space straight x space 10 cubed end fraction
space 494 space straight x space 10 to the power of negative 9 end exponent space straight m equals space 494 space nm

Question
CBSEENCH11008100

The ionization energy of He+ is 19.6 x 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is

  • 4.41 x 10-16 J atom–1

  • -4.41 x 10-17 J atom–1

  • -2.2 x 10-15 J atom–1.

  • 8.82 x 10-17 J atom–1.

Solution

B.

-4.41 x 10-17 J atom–1

Ionisation energy of He+
= 19.6 × 10-18 J
E1 (for H) × Z2= IE
E1 × 4 = – 19.6 × 10-18 J
E1 (for Li2+) = E1 for H × 9
space equals space minus space fraction numerator 19.6 space straight x space 10 to the power of negative 18 end exponent over denominator 4 end fraction space straight x space 9 space space equals space minus 44.1 space straight x space 10 to the power of negative 18 end exponent space straight J
-4.41 x 10-17 J atom–1