[nicnac89]

Hey, I've been trying to do this Qu for ages and i'm sure it's pretty simple but I just can't think how to do it:

A second order reaction with an initial concentration of 8 x 10^-3 M of reactant A gives after 400s a concentration of A equal to 6.66 x 10^-3 M. Determine:
i) the rate constant for the reaction
ii) the half-life for the reaction

Any help would be much appreciated,
Nicola

[kryptoniitti]

Hey Nicola! To get the values asked here, you need to find the correct equations. Pick and choose from these. Then then problem reduces to plotting the data points and solving for the asked variables.

[nicnac89]

Hrm...we've never been shown how to solve a problem like this by plotting data points.

If i choose the equation
rate = k [A]^2

and calculate the rate as (8x10^-3) - (6.66x10^-3)/400s
                                 = 3.35 x 10^-6 mol dm^-3 s^-1

then solve for k:
3.35x10^-6 = k (8x10^-3)^2
                = k (6.4x10^-5)
        0.052 = k

...does that seem reasonable? 

[Valdorod]

you need to use the integrated form for second order kinetics

1/[A]_t = 1/[A]₀ + kt

you have the initial concentration [A]₀ = 8E-3M
you have the concentration at time t [A]_t = 6.66E-3M
and you have the time t = 400s

plug in and solve for k

then use the equation for half life for a second order reaction and solve for it.

t _1/2 = 1/([A]₀*k)

you have both the initial concentration and k from part a.

Valdo