The rate constant for the equation 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.

The concentration of C2F4 units of M is 0.0021 M.

The molarity of the solution of C2F4 is obtained from;

Molarity = Number of moles/volume

Number of moles = 0.105 mol

Volume =  4.00 L

Molarity = 0.105 mol/ 4.00 L = 0.026 M

Recall that the reaction is second order hence;

1/[A] = kt + 1/[Ao]

[A]  = concentration of  C2F4 at time = t

[Ao] = Initial concentration of C2F4

k = rate constant

t = time taken

1/[A] = (0.0410 × 10800) + 1/ 0.026

[A] = 0.0021 M

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0.002079 M

Explanation:

2 C2F4 → C4F8

Using the differential rate equation, we have;

-d[A] / dt = k x [A]²

Where [A] represent concentration of reactant; C2H4.

Upon collecting like terms we have;

(1 / [A]²) d[A] = -k x dt

Integrating both sides;

∫(1 / [A]²) d[A] = ∫-k dt

This leaves us with;

-1/[A] + 1 / [Ao] = -k ∆t

Where [A] = Final Concentration, [Ao] = Initial concentration

Arranging the equation gives us;

1 / [A] = 1 / [Ao] + k ∆t

Time = 3 hours = 10800 s (upon converting to seconds)

Inserting the values, we have;

1 / [A] = 1 / (0.105 moles / 4 L) + (0.0410 M−1 s −1 ) x (10800s)

1 / [A] =  = 38.10 + 442.8 = 480.9

[A] = 0.002079 M

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