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For the reaction, N_{2}+3H_{2}O\rightarrow 2NH_{3}, \: \frac{d\left [ NH_{3} \right ]}{dt}=2\times 10^{-4} mol L^{-1}s^{-1}, the value of \frac{-d\left [ H_{2} \right ]}{dt} would be:

  • Option 1)

    4 x 10-4 mol L-1 s-1

  • Option 2)

    6 x 10-4 mol L-1 s-1

  • Option 3)

    1 x 10-4 mol L-1 s-1

  • Option 4)

    3 x 10-4 mol L-1 s-1

 

Answers (1)

best_answer

 

Rates in presence of stoichiometry of reactants/products -

When stoichiometry coefficients of reactants/ products are not equal to one, the rate of disappearance of  & the rate of appearance of products is divided by their respective stoichiometric coefficients

- wherein

e.g.\:2HI(g)\rightarrow H_{2}(g)+I_{2}(g)

r=\frac{-1}{2}.\frac{d}{dt}[HI]

=\frac{+d}{dt}[H2]=\frac{+ d}{dt}[I_{2}]

 

 Fix the reaction, it shoube be

N_{2}+3H_{2}\rightarrow 2NH_{3}

r=-\frac{d[N_{2}]}{dt}=-\frac{1}{3}\frac{d[H_{2}]}{dt}=\frac{1}{2}\frac{d[NH_{3}]}{dt}

\Rightarrow -\frac{1}{3}\frac{d[H_{2}]}{dt}=\frac{1}{2}\frac{d[NH_{3}]}{dt}

\Rightarrow \frac{d[H_{2}]}{dt}=\frac{3}{2}\frac{d[NH_{3}]}{dt}=3\times 10^{-4} Ms^{-1}

 

 


Option 1)

4 x 10-4 mol L-1 s-1

This is incorrect option

Option 2)

6 x 10-4 mol L-1 s-1

This is incorrect option

Option 3)

1 x 10-4 mol L-1 s-1

This is incorrect option

Option 4)

3 x 10-4 mol L-1 s-1

This is correct option

Posted by

prateek

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