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Delta(f)H^(ɵ) of hypothetical MX is -150...

`Delta_(f)H^(ɵ)` of hypothetical MX is `-150 kJ mol^(-1)` and for `MX_(2)` is `-600 kJ mol^(-1)`. The enthalpy of disproportionation of MX is `=-100 x kJ mol^(-1)`.
Find the value of x.

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The correct Answer is:
C
`2MX rarr M+MX_(2) Delta H^(ɵ) = ?`
`M(s) + (1)/(2) xx 2 rarr MX Delta_(f)H_(1)^(ɵ) = -150 kJ mol^(-1)`
`M(s) + X_(2) rarr MX_(2) Delta_(f)H_(2)^(ɵ) = -600 kJ mol^(-1)`
`:. Delta H^(ɵ) = Delta_(f)H_(2)^(ɵ)-2Delta_(f)H_(1)^(ɵ)`
`= -600 - (2 xx 150) = -300 kJ mol^(-1)`
`:. -100 x = -300`
`:. x = 3`
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