Homework Help: Chemistry

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?

No one has answered this question yet.

Answer this Question

Related Questions

chemistry - Nitric oxide reacts with bromine gas at elevated temperatures ...
Chemistry - 1. Nitric oxide reacts with bromine gas at elevated temperatures ...
chemistry kinetics - Decomposition of an organic compound A follows first order ...
Chemistry (rate reactions) - The rate constant for the thermal decomposition at ...
chemistry - The reaction Cl(g)+ O3(g) --> ClO(g)+ O2(g) is the first-...
chemistry - The reaction Cl(g)+ O3(g) --> ClO(g)+ O2(g) is the first-...
Physical Chemistry - At 25.0 degree celcius, the equilibrium 2 NOBr (gas) --&...
College Chemistry - The following data are obtained for the decomposition of ...
chemsitry - hi I need help with this question please. 1. consider the ...
Chemistry - Problem: In a first order decomposition reaction, 50% of a compound ...

For Further Reading

First Name:
School Subject:
Answer: