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Answer 1 · 2017-12-14T16:04:41

#DeltaHr° = -3684.2 kJmol^(-1)#

#DeltaHr° = SigmaDeltaHf°_p - SigmaDeltaHf°_r#
p = products; r = reactants

#SigmaDeltaHf°_p = (2 xx -1675.7) + (6 xx -296.8)#
#SigmaDeltaHf°_p = -5132.2 kJmol^(-1)#

#O_2# has a standard enthalpy of formation of zero, so does not need to be included in the calculation.

#SigmaDeltaHf°_r = (2 xx-724.0)#
#SigmaDeltaHf°_r = -1448kJmol^(-1)#

#DeltaHr° = (-5132.2) - (-1448)#
#DeltaHr° = -3684.2kJmol^(-1)#

Answer 2 · 2017-12-14T16:08:44

#"-3684.2 kJ/mol"#

Standard enthalpy of reaction (#ΔH_f^0#) is given by

#DeltaH_f^0 = SigmaDeltaH_"f(products)"^0 - SigmaDeltaH_"f(reactants)"^0#

#2Al_2S_(3(s)) + 9O_(2(g)) -> 2Al_2O_(3(s)) + 6SO_(2(g))#

#DeltaH_f^0 = [2DeltaH_f(Al_2O_(3(s)))^0 + 6DeltaH_f(SO_(2(g)))^0] - [2DeltaH_(f(Al_2S_(3(s))))]#

#DeltaH_f^0 = ["(2 × -1675.7 kJ/mol) + (6 × -296.8 kJ/mol)"] - ["2 × -724.0 kJ/mol"]#

#DeltaH_f^0 = underline"-3684.2 kJ/mol"#