Step 1
The reaction is written:
2 NOCl (g) <=> 2 NO (g) + Cl2 (g)
Step 2
2 NOCl (g) <=> 2 NO (g) + Cl2 (g)
Initial 1 mol/1 L = 1 M 0 0
When says: "it is decomposed 9.0 %" it means:
From the initial value of NOCl only is decomposed 9.0 % of 1 M, so:
0.09 x 1 M = 0.09 M
What remains is = 1 M - 0.09 = 0.91 M = this is the concentration in equilibrium for NOCl.
Now, what is formed on the right side:
For NO) +2x = 0.09 => x = 0.09/2 => x = 0.045 M
For Cl2) +x = 0.045 M
Therefore, concentrations on the right side, in equilibrium:
For NO) 0.09 M
For Cl2) 0.045 M
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Step 3