Question

For a reaction carried out at constant volume of 1 litre :
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$
The rate of formation of $N{H}_3\left(g\right)$ at particular temperature and pressure conditions was found out to be $3.4g/min$.
What will be the rate of consumption of $N_2\left(g\right)$ at the same temperature and pressure conditions?

• A. $3.4g/min$
• B. $1.7g/min$
• C. $2.8g/min$
• D. $1.4g/min$

Answer 1

The correct option is C $2.8g/min$

$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$

$\because \text{Molar mass of}N{H}_3=17g/mol$
$\text{Rate of formation of}N{H}_3=3.4g/min=\frac{3.4}{17}mol/min\text{Rate of formation of}N{H}_3=+\frac{d\left[N{H}_3\right]}{dt}=0.2mol/min$



$\text{Overall rate}\left(r\right)=-1\times \frac{d\left[N_2\right]}{dt}=+\frac{1}{2}\times \frac{d\left[N{H}_3\right]}{dt}$

$\text{Rate of consumption of}{N}_2=-\frac{d\left[N_2\right]}{dt}=\frac{1}{2}\times 0.2mol/min\text{Rate of consumption of}{N}_2=0.1mol/min$

$\text{Molar mass of}{N}_2=28g/mol$
$\text{Rate of consumption of}{N}_2=(0.1\times 28)g/min=2.8g/min$