For a reaction N2+3H2→2NH3  , the rate of formation of ammonia was found to be 2.0 x 10- 4 mol dm-3 s-1 . The rate of consumption of H2 will be

# 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

1. A

$1.0×{10}^{-4}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$

2. B

$2.0×{10}^{-4}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$

3. C

$3.0×{10}^{-4}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$

4. D

$4.0×{10}^{-4}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$

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### Solution:

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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