Q10.92CP - Chemistry: Molecular Nature Of Matter And Change

Question

When gaseous sulfur trioxide is dissolved in concentrated sulfuric acid, disulfuric acid forms:   ${\mathbf{SO}}_3\mathbf{+}{\mathbf{H}}_2{\mathbf{SO}}_4\to {\mathbf{H}}_2{\mathbf{SO}}_7$

Use bond energies (Table 9.2, p. 353) to determine $\mathbf{\Delta H}{\mathbf{°}}_{\mathrm{rxn}}$. (The S atoms in disulfuric acid are bonded through an O atom. Assume Lewis structures with zero formal charges; BE of S-N-O is 552 kJ/mol.)

Step-by-Step Solution

Answer

The heat of reaction is 22 kJ

Step 1: Reaction

${\mathbf{SO}}_3\mathbf{+}{\mathbf{H}}_2{\mathbf{SO}}_4\to {\mathbf{H}}_2{\mathbf{SO}}_7$

$\begin{array}{c}\mathrm{Energy} \mathrm{required} \mathrm{to} \mathrm{break} \mathrm{bonds}=5\left(\mathrm{S}=\mathrm{O}\right)+2\left(\mathrm{S}-\mathrm{O}\right)+2\left(\mathrm{O}-\mathrm{H}\right)\\ =5\left(552\right)+2\left(265\right)+2\left(467\right)\mathrm{kJ}\\ =4224 \mathrm{kJ}\\ \mathrm{Energy} \mathrm{released} \mathrm{when} \mathrm{bonds} \mathrm{form}=4\left(\mathrm{S}=\mathrm{O}\right)+4\left(\mathrm{S}-\mathrm{O}\right)+2\left(\mathrm{O}-\mathrm{H}\right)\\ =4\left(552\right)+4\left(265\right)+2\left(467\right) \\ =4202 \mathrm{kJ} \end{array}$

Step 2: Heat of Reaction

Therefore heat of reaction is$=\left(4224-4202\right) \mathrm{kJ}=22 \mathrm{kJ}$