Question

For `N_(2)+3H_(2)hArr 2NH_(3)`, 1 mole `N_(2)` and 3 mol `H_(2)` are at 4 atm. Equilibrium pressure is found to be 3 atm. Hence, `K_(p)` is
A. `(1)/((0.5)(0.15)^(3))`
B. `(1)/((0.5)(0.15)^(3))`
C. `(3xx3)/((0.5)(0.15)^(3))`
D. None of these

Answer 1

Correct Answer - B
`{:(,NH_(2),+,3H_(2),hArr,2NH_(3)),("Initial mol.",1,,3,,-),("Mol at eqm.",1-x,,3-3x,,2x):}`
Total no. of moles of eqm. `=1-x+3-3x+2x`
`=4-2x`
Under initial conditions,
`PV=nRT`
`4V=4RT`
`V=RT`
Uner final conditions,
`PV=nRT`
`3V=(4-2x)RT`
As `V=RT`
`2x=1` ,brgt `x=0.5`
`:. ` At eqm.
No. of moles of `NH_(3)=2x=1`
No. of moles of `N_(2)=1-x=0.5`
No. of molesof `H_(2)=3(1-x)=1.5`
`p_(NH_(3))=x_(NH_(3))xxP=(1)/(3)xx3=1`
`p_(N_(2))=x_(N_(2))xxP=(0.5)/(3)xx3=0.5`
`p_(H_(2))=x_(H_(2))xxP=(1.5)/(3)xx3=1.5`
`:.K_(p)=((p_(NH_(3)))^(2))/((p_(N_(2)))(p_(H_(2)))^(3))=((1)^(2))/((0.5)(1.6)^(3))=(1)/((0.5)(1.5)^(3))`