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At 46^@C, Kp for the reaction, N2O(4(g...

At `46^@C, K_p` for the reaction,
`N_2O_(4(g))

Text Solution

Verified by Experts

Let `p_(NO_2) = p `atm
hence, `p_(N_2O_4) = (0.5 - p) atm`
hence `K_p = (p_(NO_2)^2)/(p_(N_2O_4)) = (p^2)/((0.5 - p)) = 0.66`
`implies p^2 + 0.66 p - 0.33 = 0`
this gives `p_(NO_2) = 0.333`atm, `p_(N_2O_4) = 0.167 `atm
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