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  • For the reaction  <img alt="\mathrm{N_{2}+3 H_{2} \rightarrow 2 N H_{3}}, \text { if } \frac{\mathrm{d}\left[N H_{3}\right]}{\mathrm{d} t}=2 \times 10^{-4} \mathrm{\: mol\: } L^{-1} \:

For the reaction
 \mathrm{N_{2}+3 H_{2} \rightarrow 2 N H_{3}}, \text { if } \frac{\mathrm{d}\left[N H_{3}\right]}{\mathrm{d} t}=2 \times 10^{-4} \mathrm{\: mol\: } L^{-1} \: s^{-1}
the value of  -\frac{\mathrm{d}H_{2} }{\mathrm{d} t}  would be?

Option: 1

6 \times 10^{-4} \mathrm{{\: mol}\: L^{-1} \mathrm{s}^{-1}}


Option: 2

4\times 10^{-4} \mathrm{mol}\, \mathrm L^{-1} \mathrm{s}^{-1}


Option: 3

1 \times 10^{-4} \mathrm{mol\: } \mathrm L^{-1} \mathrm{s}^{-1}


Option: 4

3\times 10^{-4} \mathrm{mol}\: \mathrm L^{-1} \mathrm{s}^{-1}


Answers (1)

best_answer

 \mathrm{Rate = -\frac{\mathrm{d} \left [ N_{2} \right ]}{\mathrm{d} t}= -\frac{1}{3}\frac{\mathrm{d} \left [ H_{2} \right ]}{\mathrm{d} t}= \frac{1}{2}\frac{\mathrm{d} \left [ NH_{3} \right ]}{\mathrm{d} t}}
Hence,


\frac{\mathrm{d}\left[H_{2}\right]}{\mathrm{d} t}=\frac{3}{2}\times2\times10^{-4}=3\times10^{-4}

Therefore, option(4) is correct

 

Posted by

Devendra Khairwa

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