At a certain temperature, 0.680 mol of SO3 is placed in a 3.50-L container. 2SO3 (g)...

At a certain temperature, 0.680 mol of SO3 is placed in a 3.50-L container. 2SO3 (g) <---> 2SO2(g) +O2(g) At equilibrium, 0.190 mol of O2 is present. Calculate Kc.

Solutions

Expert Solution

Here the equilibrium reaction is

initially 0.68mol of SO₃0 mole of SO₂ + 0 mole of O₂

At equilibrium

At equilibrium the moles of O₂ is x=0.136moles

SO₂=2x=0.272moles

SO₃=0.68-2x=0.68-0.272=0.408mole

Now the solution volume is 3.5L

The concentration for is

The concetration of

The concentration of

Now for the equilibrium reaction is

Answer.

queen_honey_blossom answered 1 year ago

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