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One mole of nitrogen is mixed with three...

One mole of nitrogen is mixed with three moles of hydrogen in a four litre container. If 0.25 per cent of nitrogen is converted to ammonia by the following reaction
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`, then
calculate the equilibrium constant, `K_(c)` in concentration, units.
What will be the value of `K_(c)` for the following equilibrium?
`1/2 N_(2)(g)+3/2H_(2)(g)hArrNH_(3)(g)`

Text Solution

Verified by Experts

`N_(2) + 3H_(2) rarr 2NH_(3)`
`[N_(2)] = 0.75/4, [H_(2)] = 2.25/4, [NH_(3)] = 0.50/4`
`K_(c) = [NH_(3)]^(2)/([N_(2)][H_(2)]^(3)) = (0.50)^(2)/((0.75)(2.25)^(3)) xx 16 = 0.468 L^(2)mol^(-2)`
Also for : `1/2 N_(2) + 3/2H_(2) rarr NH_(3)`
`K'c = sqrt(Kc) = 0.68`
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