Question

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :
$B r O_{4}^{-} \overset{1.82}{\to} B r O_{3}^{-} \overset{1.5 V}{\to} H B r O \overset{1.595}{\to} B r_{2} \overset{1.0652 V}{\to} B r^{-}$
Then the species undergoing disproportionation is

$B r O_{3}^{-}$

$B r O_{4}^{-}$

HBrO

Br₂

Solution

C.

HBrO

Calculate $E_{c e l l}^{°}$ corresponding to each compound undergoing disproportionation reaction. The reaction for which $E_{c e l l}^{°}$ comes out +ve is spontaneous.

$H B r O \overset{}{\to} B r_{2} E^{o} = 1.595 V, S R P (c a t h o d e) H B r O \overset{}{\to} B r O_{3}^{-} E^{o} = - 1.5 V, S O P (a n o d e) 2 H B r O \overset{}{\to} B r_{2} + B r O_{3}^{-} E_{c e l l}^{o} = S R P (c a t h o d e) - S R P (a n o d e) = 1.595 - 1.5 = 0.095 V E_{c e l l}^{0} > 0 \Rightarrow ∆ G^{o} < 0 [s p o n \tan e o u s]$

Question

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Which one of the following conditions will favour maximum formation of the product in the reaction,
$A_{2} (g) + B_{2} (g) \overset{}{⇋} X_{2} (g) ∆_{r} H = - X k J ?$

Low temperature and high pressure

Low temperature and low pressure

High temperature and low pressure

High temperature and high pressure

Solution

A.

Low temperature and high pressure

$A_{2} (g) + B_{2} (g) \overset{}{⇋} X_{2} (g); ∆ H = - x k J$

On increasing pressure equilibrium shift in a direction where number of moles decreases i.e. forward direction.

On decreasing temperature, equilibrium shift n exothermic direction i.e., forward direction.

So high pressure and lower temperature favour maximum formation of product.

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The bond dissociation energies of X₂, Y₂ and XY are in the ratio of 1 : 0.5: 1. ΔH for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X₂ will be

200 kJ mol^-1

100 kJ mol^-1

400 kJ mol^-1

800 kJ mol^-1

Solution

D.

800 kJ mol^-1

Let B.E of x₂,y₂ and xy are x kJ mol^-1, 0.5 x kJ mol^-1 and x kJ mol^-1 respectively

$\frac{1}{2} x_{2} + \frac{1}{2} y_{2} \to x y; ∆ H = - 200 k J m o l^{- 1} ∆ H = - 200 = Σ (B . E)_{r e a c \tan t} - Σ (B . E)_{p r o d u c t} = [\frac{1}{2} x (x) + \frac{1}{2} x (0.5) x] - [1 x (x)] O n s o l v i n g w e g e t, B . E o f X_{2} = x = 800 k J m o l^{- 1}$

Question

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The correction factor ‘a’ to the ideal gas equation corresponds to

Density of the gas molecules

Volume of the gas molecules

Forces of attraction between the gas molecules

Electric field present between the gas molecules

Solution

C.

Forces of attraction between the gas molecules

In the real gas equation, $(P + \frac{a n^{2}}{V^{2}}) (V - n b) = n R T$ van der Waal's constant, 'a' signifies intermolecular forces of attraction.

Question

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Which one is a wrong statement?

Total orbital angular momentum of electron in 's' orbital is equal to zero

An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers

The value of m for d_z2 is zero

The electronic configuration of N atom is

Solution

D.

The electronic configuration of N atom is

According to Hund's Rule of maximum multiplicity, the correct electronic configuration of N-atom is

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NEET chemistry

The bond dissociation energies of X₂, Y₂ and XY are in the ratio of 1 : 0.5: 1. ΔH for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X₂ will be

200 kJ mol^-1

100 kJ mol^-1

400 kJ mol^-1

800 kJ mol^-1

The correction factor ‘a’ to the ideal gas equation corresponds to

Density of the gas molecules

Volume of the gas molecules

Forces of attraction between the gas molecules

Electric field present between the gas molecules

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Location

For Daily Free Study Material Join wiredfaculty

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Download Android App

Ncert Book Download

NEET chemistry

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :BrO4- →1.82BrO3-→1.5 VHBrO→1.595Br2→1.0652 VBr-Then the species undergoing disproportionation is BrO3- BrO4- HBrO Br2

Which one of the following conditions will favour maximum formation of the product in the reaction,A2 (g) + B2 (g) ⇋ X2 (g)∆r H = - XkJ? Low temperature and high pressure Low temperature and low pressure High temperature and low pressure High temperature and high pressure

The bond dissociation energies of X2, Y2 and XY are in the ratio of 1 : 0.5: 1. ΔH for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X2 will be 200 kJ mol-1 100 kJ mol-1 400 kJ mol-1 800 kJ mol-1

The correction factor ‘a’ to the ideal gas equation corresponds to Density of the gas molecules Volume of the gas molecules Forces of attraction between the gas molecules Electric field present between the gas molecules

Which one is a wrong statement? Total orbital angular momentum of electron in 's' orbital is equal to zero An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers The value of m for dz2 is zero The electronic configuration of N atom is

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The bond dissociation energies of X₂, Y₂ and XY are in the ratio of 1 : 0.5: 1. ΔH for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X₂ will be

200 kJ mol^-1

100 kJ mol^-1

400 kJ mol^-1

800 kJ mol^-1

The correction factor ‘a’ to the ideal gas equation corresponds to

Density of the gas molecules

Volume of the gas molecules

Forces of attraction between the gas molecules

Electric field present between the gas molecules