When 1 mole of N2 and 1 mole of H2 is enclosed in 5L vessel and the reaction is allowed to attain equilibrium, it is found that at equilibrium there is x mole of H2 . The number of moles of NH3 formed would be :-

Question

When 1 mole of N2 and 1 mole of H2 is enclosed in 5L vessel and the reaction is allowed to attain equilibrium, it is found that at equilibrium there is x mole of H2 . The number of moles of NH3 formed would be :- (A) (2 x/3) (B) (2 (1 + x)/3) (C) (2 (1 - x)/3) (D) ((1 - x)/2)

Tardigrade Admin · Accepted Answer

N2+3H2leftharpoons2NH3110 t=0⇒ 1 mol 1 mol 0 Eq. ⇒ (1 - y)(1 - 3 y)2y Given (1-3 y)=x(. moles of .H 2) y=(1 - x/3) ∴ Moles of (NH)3=2y=(2 (1 - x)/3)

Q. When $1$ mole of $N_{2}$ and $1$ mole of $H_{2}$ is enclosed in $5 L$ vessel and the reaction is allowed to attain equilibrium, it is found that at equilibrium there is $x$ mole of $H_{2}$ . The number of moles of $N H_{3}$ formed would be :-

$\frac{2 x}{3}$

$\frac{2 ( 1 + x )}{3}$

$\frac{2 ( 1 - x )}{3}$

$\frac{( 1 - x )}{2}$

$N_{2} + 3 H_{2} ⇌ 2 N H_{3} 110$
$t = 0 \Rightarrow$ $1$ mol $1$ mol $0$
Eq. $\Rightarrow (1 - y) (1 - 3 y) 2 y$
Given $(1 - 3 y) = x ($ moles of $H_{2})$
$y = \frac{1 - x}{3}$
$∴$ Moles of $(N H)_{3} = 2 y = \frac{2 ( 1 - x )}{3}$

Q. When 1 mole of N2​ and 1 mole of H2​ is enclosed in 5L vessel and the reaction is allowed to attain equilibrium, it is found that at equilibrium there is x mole of H2​ . The number of moles of NH3​ formed would be :-

32x​

32(1+x)​

32(1−x)​

2(1−x)​