D.
A more accurate answer would be -50 m/s. 50 m/s is the change in speed.
Answer : The activation energy for the reaction is, 119.7 J
Explanation :
According to the Arrhenius equation,

or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 271 K
= rate constant at 281 K = 
= activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 271 K
= final temperature = 281 K
Now put all the given values in this formula, we get:
![\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{271K}-\frac{1}{281K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B2K_1%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B271K%7D-%5Cfrac%7B1%7D%7B281K%7D%5D)

Therefore, the activation energy for the reaction is, 119.7 J
Solution
Given mass of FeBr2 = 85.5 g
Molar mass of FeBr2 = 215.65
No. of Moles of FeBr2 = Given Mass / Molar mass = 85.5 / 215.65
No. of Moles = 0.3964
Given Volume is 450 mL i.e. 0.450 L
Now Concentration = 0.3964/0.450
Hence Concentration = 0.881 mol/L