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?
To determine the value of 1 divided by the concentration of A, we need to consider the stoichiometry of the reaction. From the given information, we can assume the following balanced chemical equation for the decomposition of NOBr:
2NOBr(g) → 2NO(g) + Br2(g)
Since the rate equation involves the concentration of A (NOBr), we need to determine its stoichiometric coefficient. In this case, the coefficient is 2, indicating that two molecules of NOBr decompose to yield two molecules of NO and one molecule of Br2.
Now, let's consider the initial concentration of NOBr, which is given as 0.0680 moles per liter. To find the value of 1 divided by the initial concentration of NOBr, we can use the expression:
1 divided by concentration of A(initially) = 1 divided by (0.0680 mol/L)
Calculating this expression gives us the value of 1 divided by the initial concentration of NOBr.
By substituting the values of k = 0.800 1 divided by MS, the initial concentration, and the desired percent decomposition (75%), into the integrated rate law equation, we can solve for the time (t) it takes for 75% of the sample to decompose.