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](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BAB%5D%7D%20%3D%20%5Cfrac%7B1%7D%7B%5BAB_%7B0%7D%20%5D%7D%20%2B%20kT%20%5C%5C%20%5C%5C%20%5Cfrac%7B1%7D%7B%5BAB%5D%7D%20%3D%20%5Cfrac%7B1%7D%7B0.210%7D%20%2B%20%285.5%2A%2010%5E%7B-2%7D%29%2A70%20sec%20%5C%5C%20%5C%5C%20%5C%5C%20%5C%5BAB%5D%20%3D%200.116)
<span>Now, Since at 70 s, [AB] = 0.116 M,
then amount of AB lost: </span>
<span>0.210 M - 0.116 M = 0.094 M
</span>
Now, according to the stoichiometry of the reaction,
<span>AB : A : B = 1 : 1 : 1,
</span>
<span>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.</span>
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,
<span>Now, Since at 70 s, [AB] = 0.116 M,
then amount of AB lost: </span>
<span>0.210 M - 0.116 M = 0.094 M
</span>
Now, according to the stoichiometry of the reaction,
<span>AB : A : B = 1 : 1 : 1,
</span>
<span>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.</span>