The rate law for the reacion, 2NO +Cl2 ---> 2NOCL is: Rate = k [NO]^2 [CL2] A re

Question

The rate law for the reacion, 2NO +Cl2 ---> 2NOCL is:

Rate = k [NO]^2 [CL2]

A reaction is performed by mixing .020 mol of NO with .020 mol of Cl2 in a 1.0 L container. How many times faster is the reaction initially than at the time when half of the NO has been consumed?

PLEASE NOTE: this is how far I've gotten so far...'

initial rate = k [.02]^2 [.02]

rate 1/2 = k [.01]^2 [.015]

I am not sure why I am putting in a concentration of .015 though for Cl2...can someone help break this down for me? Thanks!

Explanation / Answer

Rate = k[NO]^2[Cl2]

Initially [NO] = 0.020 M and [Cl2] = 0.020 M

Rate1 = k x 0.020^2 x 0.020 = 8.0 x 10^(-6)k M/s


.....2 NO...+...Cl2...=>...2 NOCl

I...0.020.......0.020..............0

C....-2a...........-a...............+2a

F..0.020-2a...0.020-a..........2a


When half the NO is consumed:

[NO] = 0.020 - 2a = 0.020/2 = 0.010 M

=> a = 0.005

[Cl2] = 0.020 - a = 0.015 M


Rate2 = k x 0.010^2 x 0.015 = 1.5 x 10^(-6)k M/s


Rate1/Rate2 = 8.0 x 10^(-6)k/1.5 x 10^(-6)k = 5.3

Thus the iintial rate is 5.3 times faster

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