Answer:
[tex][A]=0.0026M[/tex]
Explanation:
Hello.
In this case, since the second-order rate law is:
[tex]\frac{dC_A}{dt}=-kC_A^2[/tex]
Whereas the subscript A accounts for C2F4 and its integration turns out into:
[tex]\frac{1}{[A]}= \frac{1}{[A_0]} +kt[/tex]
Thus, for the initial concentration of C2F4 computed via the 0.1 mol in the 2.31-L container:
[tex][A]_0=\frac{0.1mol}{2.31L} =0.043M[/tex]
The final concentration after 2.1 h is:
[tex]\frac{1}{[A]}= \frac{1}{0.043M} +\frac{0.048}{M*s} *\frac{3600s}{1h}*2.1h\\\\\frac{1}{[A]}=\frac{386.1}{M}[/tex]
Solving for the final concentration of C2F4, we obtain:
[tex][A]=\frac{M}{386.1} =0.0026M[/tex]
Best regards.
