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The gas phase decomposition of hydrogen peroxide at 400 °C

H2O2(g)H2O(g) + ½ O2(g)

is second order in H2O2 with a rate constant of 0.650 M-1 s-1.

If the initial concentration of H2O2 is 8.00×10-2 M, the concentration of H2O2 will be 1.70×10-2 M after
seconds have passed.

Respuesta :

sebassandin

Answer:

t = 71.3 s

Explanation:

Hello there!

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

[tex]\frac{1}{[H_2O_2]} =\frac{1}{[H_2O_2]_0}+kt[/tex]

Thus, given the initial and final concentration of hydrogen peroxide and the rate constant, we obtain the following time:

[tex]\frac{1}{[0.0170M]}-\frac{1}{0.0800M}=0.650M^{-1}s^{-1}t\\\\t=\frac{46.32M^{-1}}{0.650M^{-1}s^{-1}} \\\\t=71.3s[/tex]

Best regards!

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