Question

I’m confused on how to do thisPSA: the writing in the boxes are not a part of the question

Answer 1

The reaction is written as:

2 NOCl (g) <=> 2 NO (g) + Cl2 (g)

Initial concentration of NOCl = 1 mol/ 1 L = 1 M

9.0 % is decomposed => 0.09 x 1 M = 0.09 M

It remains (equilibirum) = 1 M - 0.09 M= 0.91 M

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Inicial concentration of NO and Cl2 is equal to 0

In equilibrium:

For NO) +2x = 0.09 M (is formed) => x = 0.045 M

For Cl2) +x = 0.045 M

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Equilibrium concentrations:

For NOCl = 0.91 M

For NO = 0.09 M

For Cl2 = 0.045 M

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K is calculated as (Kc):

Kc =

`Kc\text{ = }(\lbrack NO\rbrack^2\lbrack Cl2\rbrack)/(\lbrack NOCl\rbrack^2)=((0.09M)^2(0.045M))/((0.91)^2)=4.40x10^(-4)`

Remember: for the next reaction: a A + b B <=> c C + d D:

`Kc\text{ = }(\lbrack C\rbrack^c\lbrack D\rbrack^d)/(\lbrack A\rbrack^a\lbrack B\rbrack^b)`

Answer: K = Kc = 4.40x10^-4