MCQ
$0.6$ $mole$ of $NH_3$ in a reaction vessel of $2\,dm^3$ capacity was brought to equilibrium. The vessel was then found to contain $0.15\, mole$ of $H_2$ formed by the reaction $2N{H_{3(g)}} = {N_{2(g)}} + 3{H_{2(g)}}$ Which of the following statements is true
  • A
    $0.15 $ $mole$ of the original $NH_3$ had dissociated at equilibrium
  • B
    $0.55$ $mole$ of ammonia is left in the vessel
  • C
    At equilibrium the vessel contained $0.45$ $mole$ of ${N_2}$
  • The concentration of $NH_3$ at equilibrium is $0.25$ $mole$ per $dm^3$

Answer

Correct option: D.
The concentration of $NH_3$ at equilibrium is $0.25$ $mole$ per $dm^3$
d
$2 NH _3\quad\quad \rightleftharpoons \quad\quad \quad N _2+3 H _2( g )$

$\quad (2) \quad\quad\quad\quad\quad\quad\quad (2)$

$\text { initially } 0.6\quad \quad\quad\quad\quad 0 \quad\quad\quad 0$

$\text { equation } 0.6-2 \quad\quad x \quad\quad\quad\quad 3x$

but $3 x =0.15$

$\left[ NH _3\right]=0.6-2 x =0.6-2 \times 0.05$

$=0.5$

per $dm ^3=\frac{0.5}{2}=0.25\, mol / dm ^3$

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