At 450C the equilibrium constant KP for the reaction N2+3H2⇌2NH3 was found to be 1.6×10−5 at a pressure of 200 atm. If N2 and H2 are taken in a ratio 1:3. What is the % of NH3 formed at this temperature?
A
9
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B
18
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C
36
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D
24
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Solution
The correct option is B 18 N2+3H2⇌2NH3 1 3 0 Moles before equation (1−x)3(1−x)2x Moles after equation KP=(2x2)(4−2x)2(1−x)[3(1−x)]3×P2 KP=16×x2×(2−x)227(1−x)4×P2 or1.6×10−5=16x2×(2−x)227(1−x)4×P2 x2(2−x)2(1−x)4=1.6×10−5×27×(200)216 =1.6×10−6×27×(200)216 ∴x(2−x)(1−x)2=200×10−3×√27=1.039 ∴x=0.301 ∴MoleofNH3formed= 2×0.301=0.602 Total moles at equilibrium = 4−2×0.301=3.398 ∴ % of NH3 at equilibrium=0.6023.398×100=17.76