Question

If the initial concentration of ab is 0.210 m , and the reaction mixture initially contains no products, what are the concentrations of a and b after 70 s ?

Answer 1

ANs: [A] = [B] = 0.094 M

The following reaction was monitored as a function of time: 
AB --> A + B 
A plot of 1/[AB] versus time is a straight line with slope, K = 5.5×10^−2 M * s. 

Now, 

$\frac{1}{[AB]} = \frac{1}{[AB_{0} ]} + kT \\ \\ \frac{1}{[AB]} = \frac{1}{0.210} + (5.5* 10^{-2})*70 sec \\ \\ \\ \[AB] = 0.116$

Now, Since at 70 s, [AB] = 0.116 M,

then amount of AB lost: 
0.210 M - 0.116 M = 0.094 M 

Now, according to the stoichiometry of the reaction, 
AB : A : B = 1 : 1 : 1, 

so both [A] and [B] gained the same number of moles and thus have same concentration, as [AB] lost.

So, [A] = [B] = 0.094 M after 70 s.