InterviewSolution
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InterviewSolution
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one mole of HI is introduced into a vessel held at constant temperature . when equilibrium is reached it is found that 0.1mole of i2 have been formed . find kc |
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Answer» One MOLE of H 2
, 2 moles of I 2
and 3 moles of HI are injected in a one litre flask. x moles of H 2
will react with x moles of I 2
to form 2X moles of HI. Total number of moles of HI present at equilibrium will be 3+2x 1−x moles of H 2
and 2−x moles of I 2
will remain at equilibrium.
The equilibrium constant K c
= [H 2
][I 2
] [HI] 2
45.9= [ 1 L 1-x mol
][ 1 L 1-x mol
] [ 1 L 3+2x mol
] 2
45.9= [1−x][2−x] [3+2x] 2
45.9(1[2−x]−x[2−x])=3[3+2x]+2x[3+2x] 45.9(2−x−2x+x 2 )=9+6x+6x+4x 2
91.8−137.7x+45.9x 2 =9+12x+4x 2
41.9x 2 −149.7x+82.8=0 This is quadratic equation with solution x= 2a −b± b 2 −4ac
x= 2(41.9) −(−149.7)± (−149.7) 2 −4(41.9)(82.8)
x= 83.8 149.7±92.4
x= 83.8 149.7±92.4
x=2.88 or x=0.684 The value x=2.88 is discarded as it will lead to negative value of number of moles. Hence, x=0.684 The equilibrium concentrations are [HI]=3+2x=3+2(0.684)=4.368 mol/L [H 2
]=1−x=1−0.684=0.316 mol/L [I 2
]=2−x=2−0.684=1.316 mol/L Explanation: i think this is correct answer |
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