Methanol liquid burns readily in air. One way to represent this equilibrium is: CH3OH(l) + 3/2...

Methanol liquid burns readily in air.

One way to represent this equilibrium is:
CH₃OH(l) + 3/2 O₂(g) <-------->CO₂(g) + 2 H₂O(g

We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.

1)	2 CO₂(g) + 4 H₂O(g)	<----->	2 CH₃OH(l) + 3 O₂(g)	K₁ =
2)	CO₂(g) + 2 H₂O(g)	<------>	CH₃OH(l) + 3/2 O₂(g)	K₂ =
3)	2 CH₃OH(l) + 3 O₂(g)	<----->	2 CO₂(g) + 4 H₂O(g)	K₃ =

Expert Solution

queen_honey_blossom answered 2 years ago

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