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The gas phase decomposition of nitrogen dioxide at 383 °C

NO2(g)NO(g) + ½ O2(g)

is second order in NO2 with a rate constant of 0.540 M-1 s-1.
If the initial concentration of NO2 is 0.477 M, the concentration of NO2 will be
M after 12.4 seconds have passed.

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Answer:

[tex][NO_2]=0.112M[/tex]

Explanation:

Hello there!

In this case, since the second-order integrated law is given by the following equation:

[tex]\frac{1}{[NO_2]} =\frac{1}{[NO_2]_0}+kt[/tex]

Thus, given the initial concentration of the nitrogen dioxide gas, the rate constant and the elapsed time, we obtain:

[tex]\frac{1}{[NO_2]}= \frac{1}{0.477M} +0.54M^{-1}s^{-1}\\\\\frac{1}{[NO_2]}=8.933M^{-1}[/tex]

[tex][NO_2]=\frac{1}{8.933M^{-1}} =0.112M[/tex]

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