Question

For the chemical reaction,

\[\ce{N2_{(g)} + 3H2_{(g)} <=> 2NH3_{(g)}}\]

The correct option is:

Options

• \[\ce{-\frac{1}{3} \frac{d[H2]}{dt} = -\frac{1}{2} \frac{d[NH3]}{dt}}\]

• \[\ce{-\frac{d[N2]}{dt} = 2 \frac{d[NH3]}{dt}}\]

• \[\ce{-\frac{d[N2]}{dt} = \frac{1}{2} \frac{d[NH3]}{dt}}\]

• \[\ce{3 \frac{d[H2]}{dt} = 2 \frac{d[NH3]}{dt}}\]

Answer 1

\[\ce{-\frac{d[N2]}{dt} = \frac{1}{2} \frac{d[NH3]}{dt}}\]

Explanation:

The given reaction is

\[\ce{N2_{(g)} + 3H2_{(g)} <=> 2NH3_{(g)}}\]

The rate is given as

Rate = \[\ce{- \frac{1}{a} \frac{d[A]}{dt} = - \frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt}}\]

Here:

a = 1 for N_2​,

b = 3 for H₂​,

c = 2 for NH₃,

So, 

Rate = \[\ce{- \frac{1}{1} \frac{d[N2]}{dt} = -\frac{1}{3} \frac{d[H2]}{dt} = \frac{1}{2} \frac{d[NH3]}{dt}}\]