Questions

# For a reaction ${\mathrm{N}}_{2}+3{\mathrm{H}}_{2}\to 2{\mathrm{NH}}_{3}$  , the rate of formation of ammonia was found to be  . The rate of consumption of ${\mathrm{H}}_{2}$ will be

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a
$1.0×{10}^{-4}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$
b
$2.0×{10}^{-4}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$
c
$3.0×{10}^{-4}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$
d
$4.0×{10}^{-4}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$
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detailed solution

Correct option is C

We will have $-\frac{\mathrm{d}\left[{\mathrm{N}}_{2}\right]}{\mathrm{d}t}=-\frac{1}{3}\frac{\mathrm{d}\left[{\mathrm{H}}_{2}\right]}{\mathrm{d}t}=\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{NH}}_{3}\right]}{\mathrm{d}t}$

It is given that $\frac{\mathrm{d}\left[{\mathrm{NH}}_{3}\right]}{\mathrm{d}t}=2×{10}^{-4}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$ . Hence $-\frac{\mathrm{d}\left[{\mathrm{H}}_{2}\right]}{\mathrm{d}t}=\frac{3}{2}\frac{\mathrm{d}\left[{\mathrm{NH}}_{3}\right]}{\mathrm{d}t}=3×{10}^{-4}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$

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