Answer:
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
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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