chem-rate of reaction

Thereaction between heated sodium and halogen vapour:
2Na(s) + F2(g) → 2NaF(s)
2Na(s) + Cl2(g) → 2NaCl(s)
2Na(s) + Br2(g) → 2NaBr(s)
2Na(s) + I2(g) → 2NaI(s)

The above equations can be generalized as:
2Na(s) + X2(g) → 2NaX(s)

Such reaction can be divided into steps:
(1) 2Na(s) → 2Na(g)
(2) X-X(g) → 2X(g)
(3) 2Na(g) → Na^+(g) + e^-(g)
(4) 2X(g) + 2e^- → 2X^-(g)
(5) Na^+(g) + X^-(g) → NaX(g)

Factors affecting reaction rates:

1st Factor :
Among the 4 halogens, the atom of I is the biggest in size. In step (2), since theI-I bond has the longest bond length and thus the I-I bond is most easily tobreak. Considering only this factor, iodine has the fastest rate.

2nd Factor :
In step (4), since I atom is the biggest in size among the 4 halogens and thusit forms cations least readily. This is because the attraction of nucleus of Itowards the incoming electron is the weakest due to the longest distancebetween the nucleus and the incoming electron. Considering only this factor, iodinehas the slowest rate.

3nd Factor :
Among the 4 halide ions, I^- ion is the biggest in size. Therefore, among the 4sodium halides, the lattice the of NaI is the least stable due to the longestinterionic distance. Considering only thus factor, iodine has the slowest rate.

The combined effect and the 2nd and 3rd factors outweighs the 1st factor.Therefore, iodine has the slowest rate.

The enthalpy differences increase in the order F<Cl<Br<I , therefore the rate of reaction is F>Cl>Br>I .



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