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  • For a reaction carried out at constant volume of 1 litre : <img alt="\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})" src="https://entrancecorn

For a reaction carried out at constant volume of 1 litre :
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})
The rate of formation of  \mathrm{NH}_3(\mathrm{~g}) at particular temperature and pressure conditions was found out to be  3.4 \mathrm{~g} / \mathrm{min}
What will be the rate of consumption of  \mathrm{N}_2(\mathrm{~g}) at the same temperature and pressure conditions?

Option: 1

3.4 g/min


Option: 2

1.7 g/min


Option: 3

2.8 g/min


Option: 4

1.4 g/min


Answers (1)

best_answer

\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightarrow 2 \mathrm{NH}_3(g)
\because  Molar mass of     \mathrm{NH}_3=17 \mathrm{~g} / \mathrm{mol}
Rate of formation of  \mathrm{NH}_3=3.4 \mathrm{~g} / \mathrm{min}=\frac{3.4}{17} \mathrm{~mol} / \mathrm{min}
Rate of formation of  \mathrm{NH}_3=+\frac{d\left[\mathrm{NH}_3\right]}{d t}=0.2 \mathrm{~mol} / \mathrm{min}
            Overall rate     (r)=-1 \times \frac{d\left[N_2\right]}{d t}=+\frac{1}{2} \times \frac{d\left[N H_3\right]}{d t}

Rate of consumption of  \mathrm{N_2=-\frac{d\left[N_2\right]}{d t}=\frac{1}{2} \times 0.2 \mathrm{~mol} / \mathrm{min}}
Rate of consumption of  \mathrm{N}_2=0.1 \mathrm{~mol} / \mathrm{min}
         Molar mass of       \mathrm{N_2=28 \mathrm{~g} / \mathrm{mol}}
Rate of consumption of \mathrm{N_2=(0.1 \times 28) \mathrm{g} / \mathrm{min}=2.8 \mathrm{~g} / \mathrm{min}}.

Posted by

Ajit Kumar Dubey

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