Answer
$r=0.04\ L/mol.s\cdot [C_2F_4]^2.$
Work Step by Step
For a second-order reaction, $1/[R]=kt+1/[R]_0$, and since this graph is a straight line for $[C_2F_4]$, this is a second-order reaction, and the rate constant is $+0.04\ L/mol.s$
The rate law is then:
$r=0.04\ L/mol.s\cdot [C_2F_4]^2.$
