Question

If there is initially 2.55 M of Cl2, solve for the amount of Cl2 that is present after 10 seconds.

Please use the reaction: CH3Cl(g) + 3 Cl2(g) → CCl4(g) + 3 HCl(g) and experimental evidence:

[CH3Cl	[Cl2] (M)	Initial Rate (M/s)
0.050	0.050	0.014
0.100	0.050	0.014
0.100	0.200	0.224

(Answers Given):

0.596 M   

56.4 M

0.0177 M

1.68 M

Answer 1

Fromt the data given in the table, it is clear that,

Change in concentration of CH₃Cl does not change the initial rate

But, on increasing the concentration of Cl₂ by 4 times, the initial rate i changed by 16 times.

It indicates the threaction is zero order with respect ot CH₃Cl and second order with respect to Cl₂.

SO, overall order of reaction is = 2

And by considering the first case,

rate = k[Cl₂]²

0.014 = k (0.050)²

k = 5.6 M^-1.s^-1

And

second order rate constant integrated equation is,

1 / [Cl₂] = 1/[Cl₂]₀ + kt

1 / [Cl₂] = ( 1 / 2.55) + 5.6 ( 10)

1 / [Cl₂] = 56.4

[Cl₂] = 0.0177 M