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Calculate the equilibrium constant for the reaction : `NO(g) + 1/2 O_(2)(g) iff NO_(2)(g)`
Given, `Delta_(f)H^(@) at 298K : NO(g) = 90.4 kJ "mol"^(-1), NO_(2)(g) = 33.8 kJ "mol"^(-1)` and `DeltaS^(@)` at `298 K = -70.8 J K^(-1) "mol"^(-1)`, `R = 8.31 J K^(-1) "mol"^(-1)`.

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Verified by Experts

The correct Answer is:
`1.679xx 10^(6)`

`Delta_(r)H^(@) = Delta_(f) H^(@) ( NO_(2)) -[Delta_(f) H^(@) (NO)+(1)/(2) Delta_(f) H^(@) (O_(2))]= 33.8 -[90.4 + 0] = - 56.6kJ mol^(-1)`
`= - 56600 J mol^(-1) - 298 K ( - 70.8 JK^(-1) mol^(-1)) = - 35501.6 J mol^(-1)`
`:. - 35501.6 = - 2.303 xx 8.31 xx298 log K ` or `log K = 6.2250` or `K = 1.679 xx 10^(6)`
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