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Calculate the heat change in the reactio...

Calculate the heat change in the reaction,
`4NH_(3)(g)+3O_(2)(g)to2N_(2)(g)+6H_(2)O(l)`
at 298K given that heats of formation at 298K for `NH_(3)(g)` and `H_(2)O(l)` are `-46.0` and `-286.0kJ" "mol^(-1)` respectively. (Write answer in magnitude and KJ form)

Text Solution

Verified by Experts

The correct Answer is:
1532
`Delta_(r)H^(@)` for the reaction,
`4NH_(3)(g) + 3O_(2)(g) to 6H_(2)O(l) + 2N_(2)(g)`
`Delta_(r)H^(@) = Delta_(f)H^(@)("products") - Delta_(f)G^(@)("reactants")`
`{6Delta_(f)H^(@) [H_(2)O(l)] _ 2Delta_(f)H^(@)[N_(2)(g)]}`
`-{ 4Delta_(f) H^(@)[NH_(3)(g)] + 3Delta_(f)H^(@)[O_(2)(g)]}`
`Delta_(f)H^(@)[H_(2)O(l)] = -286.0 kJ mol^(-1)`
`Delta_(f)H^(@)[NH_(3)(g)] = -46.0 kJ mol^(-1)`
`Delta_(f)H^(@)[O_(2)(g)] = 0 and Delta_(f)^(@)[N_(2)(g)] = 0` (by convention)
`therefore Delta_(r)H^(@) = {6 (-286) + 2(0)} - { 4(-46.0) + 3(0)}`
` = -1716 + 184 = -1532 kJ`
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