# Question #295d0

Dec 20, 2016

$\textsf{A} .$

$\textsf{\Delta {H}^{\circ} = - 217.3 \textcolor{w h i t e}{x} \text{kJ/mol}}$

$\textsf{B} .$

$\textsf{262.2 \textcolor{w h i t e}{x} k J}$ released.

#### Explanation:

$\textsf{A .}$

To do this type of question you need to construct a Hess Cycle.

Hess' Law states that the overall enthalpy change of a chemical reaction is independent of the route taken.

All the substances are made from the same elements so the elements can go on the bottom.

Then construct the cycle: Note that I have divided the enthalpy of formation of $\textsf{C O}$ by 2 since the value given refers to formation of 2 moles.

Also note that the enthalpy formation of the element carbon is zero.

Hess' Law tells us that the enthalpy change of the $\textsf{\textcolor{b l u e}{b l u e}}$ route is equal to the enthalpy change of the $\textsf{\textcolor{red}{red}}$ route.

This is because the arrows start and finish in the same place.

Applying the law $\textsf{\Rightarrow}$

$\textsf{\Delta H + 0 + 106.8 = - \frac{221.0}{2}}$

$\therefore$$\textsf{\Delta H = - 110.5 - 106.8 = - 217.3 \textcolor{w h i t e}{x} k J}$

This refers to $\textsf{P {b}_{\left(s\right)} + \frac{1}{2} {O}_{2 \left(g\right)} \rightarrow P b {O}_{\left(s\right)}}$

$\textsf{B} .$

The $\textsf{{A}_{r}}$ of Pb is 207.2

So 1 mole of Pb weighs 207.2 g.

From the equation we can say:

$\textsf{207.2 \textcolor{w h i t e}{x} g \rightarrow - 217.3 \textcolor{w h i t e}{x} k J}$

$\therefore$$\textsf{1 \textcolor{w h i t e}{x} g \rightarrow - \frac{217.3}{207.2} \textcolor{w h i t e}{x} k J}$

$\therefore$$\textsf{250.0 \textcolor{w h i t e}{x} g \rightarrow - \frac{217.3}{207.2} \times 250.0 = - 262.2 \textcolor{w h i t e}{x} k J}$

The minus sign tells us that $\textsf{262.2 \textcolor{w h i t e}{x} k J}$ is released.