A reaction carried out by 1 mol of N2 and 3 mol of H2 at equilibrium. The mole fraction of NH3 as 0.012 at 500C and 10 atm pressure.What the pressure at which mole % of NH3 in equilibrium mixture is increased to 10.4?. (in atm)
A
57
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B
105
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C
200
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D
None of these
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Solution
The correct option is B105 N2⇌3H2+2NH3 1 0 0 Moles before reaction (1−x)(3−3x)2x Moles at equilibrium Given mole fraction of NH3=0.012 at P = 10 atm ∴2x(4−2x)=0.012⇒x=0.0237 ∴KP=(P′NH3)P′N2×(P′H2)3=[2x.P(4−2x)]2(1−x).P(4−2x)[(3−3x)P(4−2x)]3 KP=4x2(4−2x)2(1−x)(3−3x)2p2 =4(0.0237)2[4−2(0.0237)]2(1−0.0237)[3−3(0.0237)3]×100 KP=1.431×10−5atm−2 Let mole % of NH3 in equilibrium mixture be increased to 10.4 at pressure P. ∴2x(4−2x)=10.4100orx=0.1884 Now again using equation (i) KP=4x2(4−2x)2(1−x)(3−3x)3.P2 1.431×10−5=[4×(0.1884)2][4−2(0.1884)2][1−0.1884][3−3(0.1884)]3×P2 P=105.041atm