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For the equilibrium at 2.0 bar N(2(g)) +...

For the equilibrium at 2.0 bar `N_(2(g)) + 3H_(2(g)) harr 2NH_(3(g)) [(del)/(del p) log K_(x)]` = ?

Text Solution

Verified by Experts

The correct Answer is:
1
Kp related by `K_(x)` by equation
`Kp=K_(x)(p)^(Delta n_(g))`
`log Kp=Delta n_(g)log+log K_(x)`
`logK_(x)=logKp-Delta ng logp`
`(del)/(del p)logK_(x)=0-(Delta ng)/(p)=(0-(-2))/(2)=1`
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