Question

The second-order rate constant for the following gas-phase reaction is 0.048 1/MLaTeX: \cdot⋅s. We start with 0.1 mol C2F4 in a 2.31 liter container, with no C4F8 initially present. C2F4 LaTeX: \longrightarrow⟶ 1/2 C4F8 What will be the concentration of C4F8 after 2.1 hours?

Answer 1

Answer:

$[A]=0.0026M$

Explanation:

Hello.

In this case, since the second-order rate law is:

$\frac{dC_A}{dt}=-kC_A^2$

Whereas the subscript A accounts for C2F4 and its integration turns out into:

$\frac{1}{[A]}= \frac{1}{[A_0]} +kt$

Thus, for the initial concentration of C2F4 computed via the 0.1 mol in the 2.31-L container:

$[A]_0=\frac{0.1mol}{2.31L} =0.043M$

The final concentration after 2.1 h is:

$\frac{1}{[A]}= \frac{1}{0.043M} +\frac{0.048}{M*s} *\frac{3600s}{1h}*2.1h\\\\\frac{1}{[A]}=\frac{386.1}{M}$

Solving for the final concentration of C2F4, we obtain:

$[A]=\frac{M}{386.1} =0.0026M$

Best regards.