Find the missing ∆H(rxn)

Chemistry
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If      P4(s) + 3 O2(g) ----> P4O6(s)            ∆H(rxn) = -1640.1 kJ/mole 

and  P4(s) + 5 O2(g)----> P4O10(s)           ∆H(rxn) = -2940.1 kJ/mole

then P4O6(s) + 2 O2(g) ----> P4O10(s)     ∆H(rxn) = ?

Apr 7th, 2015

P4(s) + 3 O2(g) ----> P4O6(s)          ∆H1(rxn) = -1640.1 kJ/mole 

P4(s) + 5 O2(g)----> P4O10(s)          ∆H2(rxn) = -2940.1 kJ/mole

P4O6(s) + 2 O2(g) ----> P4O10(s)    ∆H3(rxn) = ?

∆H3(rxn) = ∆H2(rxn)-∆H1(rxn).....we are taking negative sign for ∆H1(rxn) because we need to take the enthalpy of formation for the first equation as P4O6 appears on the left hand side....

∆H3(rxn) = -2940.1 - (-1640.1) = -1300 kJ/mole (Answer)

Apr 7th, 2015

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