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The value of Kp for the equilibrium reac...

The value of Kp for the equilibrium reaction
`N_(2)O_(4)(g)

Text Solution

Verified by Experts

The correct Answer is:
71
`N_(2)O_(4)(g) Total number of moles at equilibrium
`=(1- alpha) + 2alpha`
`=(1 + alpha)`
`P_(N_(2)O_(4)) = (1-alpha)/(1+ alpha) xx P`
`K_(p) = (4 alpha^(2)P)/(1- alpha^(2))`
Given, `K_(p) =2, P= 0.5` atm
`therefore K_(p) = (4 alpha^(2)P)/(1-alpha^(2))`
`alpha=0.707=0.71`
Percentage dissociation `=0.71 xx 100 = 71`
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