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At a certain temperature this reaction follows first-order kinetics with a rate constant of 0.0197s. 2N2O5 (g) right arrow 2N2O4 (g) + O2 (g)Suppose a vessel contains N2O5 at a concentration of 0.280 M. Calculate how long it takes for the concentration of N2Os to decrease by 0.0476 M. You may assume no other reaction is important.

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Answer:

[tex]t=90.0s[/tex]

Explanation:

Hello,

In this case, for first-order kinetics we have an integrated rate law of this reaction as:

[tex]ln(\frac{[N_2O_5]}{[N_2O_5]_0} )=-kt[/tex]

Thus, we compute the time for an initial concentration of 0.280 M which ends up in 0.0476 M as shown below:

[tex]t=\frac{ln(\frac{[N_2O_5]}{[N_2O_5]_0} )}{-k}=\frac{ln(\frac{0.0476M}{0.280M})}{-0.0197s}\\ \\t=90.0s[/tex]

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