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2.0 mol of NOCl is placed in a 1.0 L flask and allowed to reach equilibrium for 2NOCl(g) <-> 2NO(g) + Cl2(g) at 30 deg C. At equilibrium the concentration of NO is found to be 0.4 mol/L. The equilibrium constant Kc, expressed as (value) * 10⁻⁴, has a value of:

1. 125 * 10⁻⁴
2. 50 * 10⁻⁴
3. 100 * 10⁻⁴
4. 12.5 * 10⁻⁴

Correct answer: 125 * 10⁻⁴

Solution

From the ICE table [NO] = 0.4, [Cl2] = 0.2, [NOCl] = 1.6, giving Kc = [NO]²[Cl2]/[NOCl]² = (0.16)(0.2)/(2.56) = 0.0125 = 125*10⁻⁴.

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