[sally125]

I have a question that I have been stumped on for hours.  I've tried putting all the values into the normal second order equation 1/[A]t = kt + 1/[A]0 but since I know the answer I know that I am doing something wrong.

I also realize that they give you the concentration of the reactant and want the concentration of the product.  Please help.

The question is:
Ammonium cyanate, NH4NCO, rearranges in water in a second-order reaction to form urea,
NH2CONH2. The (specific) rate constant for this reaction is k = 0.667 mol–1 L h–1 at 50°C.
If you begin with a solution of ammonium cyanate with a concentration of 0.500 mol L–1 at 50°C,
what concentration of urea will be produced at the end of 5 hours?

[plu]

Plug and play    What value are you getting?

[sally125]

.187 which isn't the right answer.  The answer is .313

[pantone159]

I get 3/16 as the concentration of the ammonium cyanate after 5 hours.
As you said, the concentration of the product is asked for instead...

[Donaldson Tan]

the second order equation gives you the concentration of the reactant (ammonium cyanate) at end of 5h. then use mole balance to find the corresponding concentration of the product (urea).