111/2N2 +3/2 H2 = NH3
1 3 0
1-0.2 3-0.6 0.4
K = (0.4/3)/(0.8/3)1/2*(2.4/3)3/2
K= 0.4 * 3 /√0.8 * 2.4*√2.4
K = 4*3/2√2*2.4*√2*3*4
K=5/4*2*√3
K = 5/8√3
1 mole of nitrogen is mixed with 3 mole of in a 3 litre vessel 20% of nitrogen is converted into ammonia. d equilibrium constant for d rean
1/2N2 +3/2 H2 <----> NH3
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9 Answers
11
11 mole + 3 moles
Nitrogen Hydrogen Ammonia
.8/3 m/l 2.4/3 m/l .4/3 m/l
Taking these together....
concentration of each of these is
\frac{.4/3}{(.8/3)^{1/2}\times .8^{3/2}}
Is this the answer?
10.36 L/mol is the ans.
multiply reaction by 2.
we get the equation for habers process
N2 + 3 H2 ====> 2 NH3
initial moles 1 3 0
final moles 1-0.2 3-0.6 0.4
final 0.8/3 2.4/3 0.4/3
concentration
so K=(0.4/3)2/{(0.8/3)(2.4/3)3}
so K for given reaction is= √[(0.4/3)2/{(0.8/3)(2.4/3)3}]
=0.36 l/mol