4. A quantity of 2.40moles of pure SO2Cl2gas was placed in an 8.00 L sealedflask

Question

4. A quantity of 2.40moles of pure SO2Cl2gas was placed in an 8.00 L sealedflask. At 500 K, after

equilibrium isestablished, there are 1.60 moles of the product gasSO2 present. Calculate Kc for the

reaction.SO2Cl2(g) ?? SO2(g) + Cl2 (g)

2.40 mol /8.00 L = .300M initial SO2Cl2 (g) 1.60 mol/ 8.00L = .200 M SO2 (g)at ??

         SO2Cl2 (g) ??SO2(g) + Cl2(g)

I       .300                   0          0

C      -x                     +x       +x

E       (.300-x)          .200       x

So x must = .200 soKc =(.200)2 / .100 =.400

Explanation / Answer

                                 SO2Cl2  <-----------------> SO2 +Cl2                           Since1 mole of SO2Cl2 produces 1 mole each of SO2 &Cl2   So at eqilibrium if x moles of SO2 is formed , x molesa of Cl2will be exqctly formed. AND (0.3 - x ) moles of SO2Cl2 willremain.

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---