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Bronsted Lowry Theory

For the react...

Question

For the reaction, $N_{2} + 3 H_{2} \to 2 N H_{3}, i f E_{1} a n d E_{2}$ are the equivalent masses of $N H_{3} a n d N_{2}$ respectively, then $E_{1} - E_{2}$ is:

A

1

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B

2

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C

3

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D

4

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Solution

The correct option is A 1
$N_{2} + 6 e^{-} \to 2 N^{3 -}$
$E_{2} = \frac{28}{6}, E_{1} = \frac{17}{3}$
$E_{1} - E_{2} = \frac{17}{3} - \frac{28}{6} = 1$

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Similar questions

Q. For the reaction,

N_{2} + 3 H_{2} ⟶ 2 N H_{3}

E_{1}

E_{2}

are equivalent masses of

N H_{3}

N_{2}

respectively, then

E_{1} - E_{2}

Q. If e₁ and e₂ are respectively the eccentricities of the ellipse

\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1

and the hyperbola

\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1

, then the relation between e₁ and e₂ is
(a) 3 e₁² + e₂² = 2
(b) e₁² + 2 e₂² = 3
(c) 2 e₁² + e₂² = 3
(d) e₁² + 3 e₂² = 2

Q. Equivalent masses of

N H_{3}

in the reactions are :
(i)

4 N H_{3} + 5 O_{2} ⟶ 4 N O + 6 H_{2} O

2 N H_{3} ⟶ N_{2} + 3 H_{2}

N_{2} + 3 H_{2} \to 2 N H_{3}

Molecular weight of

N H_{3}

N_{2}

x_{1}

x_{2}

respectively. Their equivalent weights are

y_{1}

y_{2}

respectively. Then

(y_{1} - y_{2})

$I f e_{1} a n d e_{2} a r e r e s p e c t i v e l y t h e e c c e n t r i c i t i e s o f t h e e l l i p s e \frac{x^{2}}{18} + \frac{y^{2}}{4} = 1 a n d t h e h y p e r b o l a \frac{x^{2}}{9} - \frac{y^{2}}{4} = 1, t h e n t h e r e l a t i o n b e t w e e n e_{1} a n d e_{2} i s$

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