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Anonymous asked in Science & MathematicsChemistry · 1 month ago

Chemistry- How would I go about solving for Kc when I don't have K?

At a particular temperature, 4.00 mole of NOCI is placed in a 1.00 L rigid container, and decomposes according to the reaction below. At equilibrium, 3.00 mol of NOCI is present. Calculate the equilibrium constant Kc for this reaction. 

                               2NOCI(g) = 2NO(g)+Cl2(g) 

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  • 1 month ago
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                            2NOCI(g)  ⇌  2NO(g)  +  Cl₂(g)     Kc

    Initial (mol/L):       4.00             0.00          0.00

    Change (mol/L):    -2y              +2y            +y

    Eqm (mol/L):   (4.00 - 2y)          2y              y

    At equilibrium:

    [NOCl] = (4.00 - 2y) mol/L = 3.00 mol/L, Hence, y = 0.500 mol/L

    [NO] = 2y mol/L = 2 × 0.500 mol/L = 1.00 mol/L

    [Cl₂] = 0.500 mol/L

    Kc = [NO]² [Cl₂] / [NOCl]² = (1.00)² (0.500) / (3.00)² = 0.0556

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