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For the reaction:
\mathrm{N}_2 \mathrm{O}_5(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_2(\mathrm{g})+1 / 2 \mathrm{O}_2(\mathrm{g})
The value of rate of disappearance of \mathrm{\mathrm{N}_2 \mathrm{O}_5} was found to be \mathrm{6.25 \times 10^{-3} \mathrm{molL}^{-1} \mathrm{~s}^{-1}}

Then choose the correct option:
 

Option: 1

The rate of formation of \mathrm{NO}_2 is \mathrm{1.25 \times 10^{-2} \mathrm{molL}^{-1} \mathrm{~s}^{-1}}.


Option: 2

The rate of formation of \mathrm{NO}_2 is 3.125 \times 10^{-3} \mathrm{molL}^{-1} \mathrm{~s}^{-1}.


Option: 3

The rate of formation of \mathrm{O}_2 is 1.25 \times 10^{-2} \mathrm{molL}^{-1} \mathrm{~s}^{-1}.


Option: 4

None of the above.


Answers (1)

best_answer

For a given reaction, if:

aA + bB \rightarrow cC + dD

Overall rate r can be expressed as:
\mathrm{ r=\frac{-1}{a} \times \frac{d[A]}{d t}=\frac{-1}{b} \times \frac{d[B]}{d t}=\frac{+1}{c} \times \frac{d[C]}{d t}=\frac{+1 d[D]}{d d t} }

where:
\mathrm{ -\frac{d[A]}{d t}= } Rate of disappearance of A
\mathrm{ -\frac{d[B]}{d t}= } Rate of disappearance of B
\mathrm{ +\frac{d[C]}{d t}= } Rate of formation of C
\mathrm{ +\frac{d[D]}{d t}= } Rate of formation of D

For:
\mathrm{ \mathrm{N}_2 \mathrm{O}_5(g) \rightarrow 2 \mathrm{NO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \\ }
\mathrm{ r=-\frac{d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=\frac{+1}{2} \times \frac{d\left[N O_2\right]}{d t}=+2 \times \frac{d\left[\mathrm{O}_2\right]}{d t} }

 

Rate of formation of \mathrm{ \mathrm{NO}_2=2 \times } Rate of disappearance of \mathrm{ \mathrm{N}_2 \mathrm{O}_5 }
Rate of formation of \mathrm{ \mathrm{NO}_2=2 \times 6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} }

Rate of formation of \mathrm{ \mathrm{O}_2=\frac{1}{2} \times } Rate of disappearance of \mathrm{ \mathrm{N}_2 \mathrm{O}_5 }
Rate of formation of \mathrm{ O_2=\frac{1}{2} \times 6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} }  

Posted by

Devendra Khairwa

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