Consider the following reversible reaction at a specific temperature: 2 NOCl ( g ) 2 NO ( g ) + Cl ₂( g ) The equilibrium constant, K_c , for this reaction is 1.8 10⁻⁵ . If the equilibrium concentrations of NOCl and Cl ₂ are 0.040 M and 0.020 M respectively, what is the equilibrium concentration of NO ?

Step-by-step solution

The equilibrium constant expression for the given reaction is: K_c = [ NO ]^2 [ Cl ₂] [ NOCl ]^2 Substituting the given equilibrium concentrations and K_c : 1.8 10⁻⁵ = [ NO ]^2 (0.020) (0.040)^2 1.8 10⁻⁵ = [ NO ]^2 (0.020) 1.6 10⁻³ Rearranging to solve for [ NO ]^2 : [ NO ]^2 = 1.8 10⁻⁵ 1.6 10⁻³ 0.020 [ NO ]^2 = 2.88 10⁻⁸ 2.0 10⁻² [ NO ]^2 = 1.44 10⁻⁶ Taking the square root on both sides: [ NO ] = 1.2 10⁻³ M Answer: 1.2 10⁻³ M