Question

2 C2F4 → C4F8 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

After three hours, concentration of C₂F₄ is 0.00208M

Step-by-step explanation:

As the units of the rate constant of the reaction are M⁻¹s⁻¹ we can know this reaction is of second order. Its integrated law is:

`(1)/([A]) =(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.

3.00 hours are in seconds:

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

Computing in the equation:

`(1)/([A]) =(1)/([0.105mol / 4L]) +0.0410M^(-1)s^(-1)*10800s\\(1)/([A]) = 480.9M^(-1)\\\`

[A] = 0.00208M

After three hours, concentration of C₂F₄ is 0.00208M