Question #b688f

2 Answers
Dec 14, 2017

DeltaHr° = -3684.2 kJmol^(-1)

Explanation:

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)

Dec 14, 2017

"-3684.2 kJ/mol"

Explanation:

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"