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Calculate the enthalpy change in the rea...

Calculate the enthalpy change in the reaction, `4NH_(3) + 3O_(2) to 2N_(2) + 6H_(2) O` at 298 K,given that the enthalpy of formation at 298 K for `NH_(3) and H_(2)O` are `- 46.0` and -`286.0 KJ mol^(-1)` respectively.

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`4NH_(3) + 3O_(2) to 2N_(2) + 6H_(2)O Delta H = ?`
Given :
`Delta H_(f)^(o) (NH_(3)) = - 46.9 kj mol^(-1)`
`Delta H)_(f)^(o) (H_(2)O) = - 286.0 kj mol^(-1)`
`Delta " "_(r)H^(o)=sumDeltaH^(o)""_(f)("Product")-sumDeltaH^(o)""_(f)("Reactant")`
` = 6 xx (-286.0) - 4 (-46.0)`
`= -1716+ 184 = - 1532 Kj mol^(-1)`
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