# Question b688f

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

$\text{-3684.2 kJ/mol}$

#### Explanation:

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

$\Delta {H}_{f}^{0} = \Sigma \Delta {H}_{\text{f(products)"^0 - SigmaDeltaH_"f(reactants)}}^{0}$

$2 A {l}_{2} {S}_{3 \left(s\right)} + 9 {O}_{2 \left(g\right)} \to 2 A {l}_{2} {O}_{3 \left(s\right)} + 6 S {O}_{2 \left(g\right)}$

$\Delta {H}_{f}^{0} = \left[2 \Delta {H}_{f} {\left(A {l}_{2} {O}_{3 \left(s\right)}\right)}^{0} + 6 \Delta {H}_{f} {\left(S {O}_{2 \left(g\right)}\right)}^{0}\right] - \left[2 \Delta {H}_{f \left(A {l}_{2} {S}_{3 \left(s\right)}\right)}\right]$

$\Delta {H}_{f}^{0} = \left[\text{(2 × -1675.7 kJ/mol) + (6 × -296.8 kJ/mol)"] - ["2 × -724.0 kJ/mol}\right]$

$\Delta {H}_{f}^{0} = \underline{\text{-3684.2 kJ/mol}}$