Question

is 0.0410 M−1 s −1 . We start with 0.105 mol C2F4 in a 4.00-liter container, with no C4F8 initially present. What will be the concentration of C2F4 after 3.00 hours ? Answer in units of M.

Answer 1

Answer:

After three hours, concentration of C₂F₄ is 0.00208M

Explanation:

The rate constant of the reaction:

2 C2F4 → C4F8 is 0.0410M⁻¹s⁻¹

As the units are M⁻¹s⁻¹, this reaction is of second order. The integrated law of a second-order reaction is:

$\frac{1}{[A]} =\frac{1}{[A]_0} +Kt$

Where [A] and [A]₀ represents initial and final concentrations of the reactant (C₂F₄), K is rate constant (0.0410M⁻¹s⁻¹) and t is time of the reaction (In seconds).

3.00 hours are in seconds:

3 hours ₓ (3600 seconds / 1 hour) = 10800 seconds

Initial concentration of C2F4 is:

0.105mol / 4.00L = 0.02625M

 

Replacing in the integrated law:

$\frac{1}{[A]_0}= \frac{1}{0.02625} +0.0410M^{-1}s^{-1}*10800s\\\frac{1}{[A]_0}=480.9M^{-1}$

[A] = 0.00208M

After three hours, concentration of C₂F₄ is 0.00208M