Question #6ea84

1 Answer
Stefan V.
Feb 3, 2015

In order to find the standard-state enthalpy for this reaction you need to substract the sum of the enthalpis of formation of the reactants from the sum of the enthalpies of formation of the products.

Each of these enthalpies of formation must be multiplied by their respective coefficients from the balanced equation. So,

#2Ag_2S_((s)) + 2H_2O_((l)) -> 4Ag_((s)) + 2H_2S_((g)) + O_(2(g))#

The standard-state enthalpy will be

#DeltaH^@ = "2 moles" * (-"20.5 kJ/mole") - ("2 moles" * (-"32.6 kJ/mole") + "2 moles" * (-"285.5 kJ/mole"))#

#DeltaH^@ = -"41.0 kJ" - (-"636.2 kJ") = "+595.2 kJ"#

Remember that the standard-state enthalpy of formation for elements is zero, that is why #Ag_((s))# and #O_(2(g))# do not appear in the calculation for #DeltaH^@#.

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