Chapter 9, Problem 9.53PAE

9.53 Using these reactions, find the standard enthalpy change for the formation of 1 mol of PhO(s) from lead metal and oxygen gas.

   $\text{PbO(s)}+\text{C}\left(\text{graphite}\right)\to \text{Pb(s)}+\text{CO(g)}$

   $\Delta \text{H}°$ = 106.8 kJ

   $\text{2C(graphite)}+\text{O2(g)}\to \text{2CO(g)}$

   $\Delta \text{H}°$= -221.0 kJ

If 250 g of lead reacts with oxygen to form lead(II) oxide, what quantity of thermal energy (in kJ) is ahsorhed or evolved?

Interpretation Introduction

Interpretation: The quantity of thermal energy absorbed or evolved has to be calculated when 250g of lead reacts with oxygen to form lead(II) oxide.

Concept introduction:

• Hess’s law states that irrespective of the multiple steps in a reaction, the total enthalpy change is the summation of all the changes.
• The Hess’s law allows us to calculate the heat of reaction for any reaction that occurs at standard conditions using heat of formation of reactants and products.
• The standard enthalpy change for a reaction can be calculated as: $\Delta H°={\displaystyle \sum n}\Delta H_f°(products)-{\displaystyle \sum n}\Delta H_f°(reac\text{tan}ts)$, where $\Delta H°$ is the enthalpy or the heat of the reaction and $\Delta H_f$ is the heat of formation.

Given:

$\begin{array}{l}Pb\text{O}(s)+\text{C}(s,graphite)\to \text{Pb}(s)\text{+ CO}(g)\Delta H°=10\text{6}\text{.8kJ (i)}\\ \text{2C}(s,graphite){\text{+O}}_{\text{2}}(g)\to 2\text{CO}(g)\Delta H°=-221.0\text{kJ}\text{(ii)}\end{array}$

Answer to Problem 9.53PAE

Solution: The quantity of thermal energy absorbed or evolved when 250g of lead reacts with oxygen to form lead(II) oxide is $\text{-262}\text{.2 kJ}$.

Explanation of Solution

The enthalpy or the standard heat of a reaction equals the difference between the sum of all the standard heats of formation of the products and the sum of all the standard heats of formation of the reactants. Here “n” is the number by which heat of formation for each reactant or product must be multiplied and we obtain it from the balanced equation.

To obtain the equation for reaction of lead with oxygen to form lead(II) oxide, subtract 2 times equation (i) from equation (ii).

$\frac{\begin{array}{l}\text{ 2C}(s,graphite){\text{+O}}_{\text{2}}(g)\to 2\text{CO}(g)\text{(ii)}\\ \text{-2PbO}(s)+2\text{C}(s,graphite)\to 2\text{Pb}(s)\text{+ 2CO}(g)\text{ (i)}\times 2\end{array}}{2\text{Pb}(s){\text{+O}}_{\text{2}}(g)\to \text{2PbO}(s)}$

$\begin{array}{l}\Delta H°={\displaystyle \sum n}\Delta H_f°(products)-{\displaystyle \sum n}\Delta H_f°(reac\text{tan}ts)\\ =\left\{\text{-221}\text{.02 - 2(106}\text{.8)}\right\}\text{kJ}\\ \text{ = - 434}\text{.62 kJ}\end{array}$

Molecular weight of lead = $\text{207}\text{.2 g/mol}$

From the balanced chemical equation, when 2 mole, i.e., $\text{2×207}\text{.2g = 414}\text{.4g}$ of lead reacted, the heat given off is $\text{- 434}\text{.62 kJ}$

Therefore, when $\text{250g}$ of lead reacted, the heat given off is

Conclusion

The quantity of thermal energy absorbed or evolved when 250g of lead reacts with oxygen to form lead(II) oxide is $\text{-262}\text{.2 kJ}$