Calculate the enthalpy change for the reaction P4O6(s)+2O2(g)=P4O10(s)?

given the following enthalpies of reaction:

P4(s)+3O2(g)=P4O6(s) Delta H = -1640.1kJ

P4(s)+5O2(g)=P4O10(s) Delta H = -2940.1kJ

I don't even know how to approach this!

1 Answer

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  • 1 decade ago
    Best Answer

    you can flip a reaction around and flip the sign for delta H

    and with hess's law you can also "add" two chemical equations and add their delta h's, for example if you have

    A + B => C + D delta H = 100

    Z + X => T + K delta H = -40

    then for

    A + B + Z + X => C + D + T + K, delta H is gonna be 100 + (-40) = 60

    and another thing, if you have the same exact thing on both sides of equation, you can cross it out for example

    Na+ + Cl- + Xe => Xe + NaCl

    is "actually" Na+ + Cl- => NaCl

    delta H doesnt change if you throw stuff away thats the same on both sides.

    so you flip the first one around you get:

    P4O6(s)=P4(s)+3O2(g) Delta H = 1640.1kJ

    you add the second one

    P4(s)+5O2(g)=P4O10(s) Delta H = -2940.1kJ

    to get

    P4O6(s)+P4(s)+5O2(g)=P4(s)+3O2(g)+P4O10(s)

    throw away the P4(s) and rewrite the 5O2(g) as 3O2(g) + 2O2(g) to get

    P4O6(s)+2O2(g)+3O2(g)=3O2(g)+P4O10(s)

    now its obvious you can also throw away the 3O2(g) to get what u needed

    P4O6(s)+2O2(g)=P4O10(s) and delta H for that is

    1640.1kJ + (-2940.1kJ) = -1300 kJ

    Source(s): school
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