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The equilibrium constant KP for CCl4(g) f15g2a33g1.jpgC(s) + 2 Cl2(g) is 0.76 at 700 K. What...

The equilibrium constant KP for CCl4(g) f15g2a33g1.jpgC(s) + 2 Cl2(g) is 0.76 at 700 K. What percentage (%) of CCl4 is converted into C and Cl2 when a flask charged with 3.00 atm of CCl4 reaches equilibrium at 700 K?

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Expert Solution

                                    CCl4 (g) <==>   C(s)    +     2 Cl2 (g)

Initial:                          3.00                                 0

change:                        -x                                 + 2x

Equilibrium:               3.00 - x                                 2x       

or, 2.28 - 0.76x = 4x2

solving the quadratic equation for x, we get:

x = 0.67 atm

Hence, equilibrium pressure of CCl4 = 3.00 - 0.67 = 2.33 atm

% of CCl4 converted into C and Cl2 is = (0.67/3.00)x100 = 22.33 %

queen_honey_blossom answered 2 years ago
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