# The reaction 2A->B is 2nd order.The initial concentration of A is 13.198M and half-life is 198 sec. How long will it take [A] to drop to 1/7 of its initial value?

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### 1 Answer

To solve this problem we need to know and use two different equations. The first is the equation for the half life of a second order reaction. It is:

`t=1/(k[A'])`

where t is the half life, k is the second order rate constant, and [A'] is the initial concentration of A. We can use this equation to calculate the rate constant k:

`198=1/(k*13.198)`

k = 3.827 L/mol*sec

We can now use the integrated rate law for a second order reaction to calculate the time required for [A] to drop to 1/7 of the initial value.

`[A]=([A'])/(1+kt[A'])`

Solving for t:

`t=(([A'])/([A]) - 1)(1/(k[A']))`

Now we can input the values and solve for t. [A'] is the initial concentration (13.198 M). [A] is 1/7 of the initial value (1.885 M). And k is the rate constant that we calculated above. Inputting all of these values into the above equation gives t=1,188.31 seconds. So the answer is that it takes 1,188.31 seconds for the concentration to drop to 1/7 of its initial value.

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