Consider the following reaction where Kc = 1.20x10-2 at 500 K: PCl; (g)PCl3 (g)

Question

Consider the following reaction where Kc = 1.20x10-2 at 500 K: PCl; (g)PCl3 (g) + Cl2 (g) A reaction mixture was found to contain 0.125 moles of PCls (2),4.99-102 moles of PCl3 (g), and 4.1 Indicate True (T) or False (F) for each of the follow I. In order to reach equilibrium PC15(g) must be produced in a 2. In order to reach equilibrium Kc must decrease 3. In order to reach equilibrium PC13 must be consumed - 4. Qc is greater than Kc. 5-The reaction is at equilibrium. No further reaction will occur.

Explanation / Answer

PCl5(g) -------------------> PCl3(g) + Cl2(g)       Kc = 1.2*10^-2

[PCl5]   = 0.125 mole/1L   = 0.125moles/L

[PCl3]   = 4.99*10^-2mole/1L   = 4.99*10^-2mole/L

[Cl2]    = 4.14*10^-2 moles/1L   = 4.14*10^-2 mole/L

    Qc   = [PCl3][Cl2]/[PCl5]

            = 4.99*10^-2 *4.14*10^-2/0.125

             = 0.0165

Qc>KC The equilibrium shift to reactant side ( reverse reaction)

1. True

2. True

3. True

4. True

5. Flase

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