Consider the following equilibrium: 2NOCl( g ) lt;—–gt; 2NO( g ) + Cl 2 ( g ) K = 1.6 X 10 -5 1.00 mole of pure NOCl and 0.958 mole of pure Cl 2…
2NOCl(g) <—–> 2NO(g) + Cl2(g) K = 1.6 X 10–5
1.00 mole of pure NOCl and 0.958 mole of pure Cl2 are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(g).
Provide all work and equations used for correct answer.