How do you calculate the standard enthalpy of reaction, $DeltaH_(rxn)^@$ for the following reaction from the given standard heats formation ( $DeltaH_f^@$ ) values: $4NH_3(g) + 5O_2(g) -> 4NO(g) + 6H_2O(g)$ ?

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Answer 1 · 2017-06-02T03:31:44

$DeltaH_"rxn"^@=DeltaH_f^@("products")-DeltaH_f^@("reactants")$

And here, $DeltaH_"rxn"^@=..................$

$(4xx90.3-6xx241.8-{4xx-45.9})*kJ*mol^-1$

$=-906*kJ*mol^-1$ .

The formation of water clearly dominates the enthalpy change, it is a thermodynamic sink.

ANd of course, $DeltaH_f^@=0$ for an element (here dixoygen) in its standard state (they even spoon-feed you this fact).

How do you calculate the standard enthalpy of reaction, DeltaH_(rxn)^@DeltaH_(rxn)^@ for the following reaction from the given standard heats formation (DeltaH_f^@DeltaH_f^@) values: 4NH_3(g) + 5O_2(g) -> 4NO(g) + 6H_2O(g)4NH_3(g) + 5O_2(g) -> 4NO(g) + 6H_2O(g)?