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$2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightleftharpoons 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$
If rate and rate constant for above reaction are $2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $3 \times 10^{-5} \mathrm{~s}^{-1}$ respectively, then calculate the concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$
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If rate and rate constant for above reaction are $2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $3 \times 10^{-5} \mathrm{~s}^{-1}$ respectively, then calculate the concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$
Solution:
2748 Upvotes
Verified Answer
The correct answer is:
0.8
The reaction is of first and for a first order reaction, rate, $R=k\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]$ $2.4 \times 10^{-5}=3 \times 10^{-5} \times\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]$
$\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=\frac{2.4 \times 10^{-5}}{3 \times 10^{-5}}=0.8 \mathrm{~mol} \mathrm{~L}^{-1}$
$\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=\frac{2.4 \times 10^{-5}}{3 \times 10^{-5}}=0.8 \mathrm{~mol} \mathrm{~L}^{-1}$
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