Question

from these data,

mathrm{s}(\text { rhombic })+mathrm{o}_{2}(g)
ightarrow mathrm{so}_{2}(g) quad delta h_{mathrm{rxn}}^{circ}=-296.06 \frac{mathrm{kj}}{mathrm{mol}}

mathrm{s}(\text { monoclinic })+mathrm{o}_{2}(g)
ightarrow mathrm{so}_{2}(g) quad delta h_{mathrm{rxn}}^{circ}=-296.36 \frac{mathrm{kj}}{mathrm{mol}}

calculate the enthalpy change for the transformation:

mathrm{s}(\text { rhombic })
ightarrow mathrm{s}(\text { monoclinic })

(monoclinic and rhombic are different allotropic forms of elemental sulfur.)

round your answer to 2 significant digits.

\frac{mathrm{kj}}{mathrm{mol}}

Explanation:

Step1: Reverse the second reaction

Reverse: $\ce{SO2(g) -> S(monoclinic) + O2(g)}$, $\Delta H^\circ_{\text{rxn}} = +296.36\ \frac{\text{kJ}}{\text{mol}}$

Step2: Add to first reaction

First reaction: $\ce{S(rhombic) + O2(g) -> SO2(g)}$, $\Delta H^\circ_{\text{rxn}} = -296.06\ \frac{\text{kJ}}{\text{mol}}$
Sum: $\ce{S(rhombic) -> S(monoclinic)}$
$\Delta H = (-296.06 + 296.36)\ \frac{\text{kJ}}{\text{mol}} = 0.30\ \frac{\text{kJ}}{\text{mol}}$

Step3: Round to 2 sig figs

$0.30\ \frac{\text{kJ}}{\text{mol}}$

Answer:

$0.30\ \frac{\text{kJ}}{\text{mol}}$