a second- order reaction of the type a + b --> p was carried out in a solution that was initially 0.075 mol dm^-3 in a and 0.050 mol dm^-3 in b. after 1.0 h the concentration of a had fallen to 0.020 mol dm^-3. a) calculate the rate constant. b) solve for the half- life of each of the reactants.
hint: answers are a) 16.2 dm^3/mol*h
b) 5.1 × 10^3 s, 2.1 × 10^3 s

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate =

If and are the inital concentrations and x the concentration reacted at time t, so and and the rate at time t is written as:

Separating variables and integrating

The integral in left side is solved by partial fractions, it can be used integral tables

Using logarithm properties (ln x - ln y = ln(x/y))

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, , , , it implies that the quantity reacted, x, is 0.03 and . Then, the value of k would be

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

but suppossing that so  , k=16.2 and the same initial concentrations. Replacing in the equation

For reactant B, so  , k=16.2 and the same initial concentrations. Replacing in the equation

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

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