Posted by **Eliz** on Monday, February 23, 2009 at 9:54pm.

Question: the rate constant for the second order decomposition of NOBr(g) is .800 1 divided by MS at 10 degree's C. Suppose a 2.50 L beaker contains NOBr at a concentration at .0680 moles per liter. How much time does it take for 75.0 percent of the sample to decompose.

I know that the equation is 1 divided by concentration of A=kt+1 divided by concentration of A(initially). I'm not sure how you get 1 divided by concentration of A. I know that k=.800 and that the initial concentration is 1 divided by(.0680) and that we are solving for t(time). How do you get the 1 divided by concentration of A with the information from the problem?

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