# Answer to Question #57767 in Other Chemistry for sofie

Question #57767

The rate constant for the equation

2 C2F4 → C4F8

is 0.0469 M−1

· s

−1

. We start with 0.203 mol

C2F4 in a 2.00-liter container, with no C4F8

initially present. What will be the concentra-

tion of C2F4 after 1.00 hour ?

2 C2F4 → C4F8

is 0.0469 M−1

· s

−1

. We start with 0.203 mol

C2F4 in a 2.00-liter container, with no C4F8

initially present. What will be the concentra-

tion of C2F4 after 1.00 hour ?

Expert's answer

For the second order reaction:

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

But according to the provided equation, the process must be described as: 2A -----> products.

Initial concentration of C2F4 is: 0.203/2=0.1015 mol/l.

1/2·A = 1/ (2 · 0.1015) + 0.0469 · 3600 =173.766

2 A = 0.0058

A =0.0115 mol/l

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

But according to the provided equation, the process must be described as: 2A -----> products.

Initial concentration of C2F4 is: 0.203/2=0.1015 mol/l.

1/2·A = 1/ (2 · 0.1015) + 0.0469 · 3600 =173.766

2 A = 0.0058

A =0.0115 mol/l

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