Chapter 10, Problem 31A

Interpretation Introduction

Interpretation: When 5.00g of propane are burned with excess oxygen the enthalpy change is to be calculated. 

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(g\right)+5{\text{O}}_{\text{2}}\left(g\right)\to 3{\text{CO}}_{\text{2}}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\Delta \text{H=}-2221\text{kJ}/\text{mol}$

Concept introduction: The term $\text{U+PV}$ is known as enthalpy of the system. It is indicated by H. It performs total energy stored in the system. Thus,

  $\text{H=U+PV}$

Here U, P and V are definite quantities.

Answer to Problem 31A

Enthalpy change is 251.86 kJ

Explanation of Solution

The given equations are as follows:

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(g\right)+5{\text{O}}_{\text{2}}\left(g\right)\to 3{\text{CO}}_{\text{2}}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\Delta \text{H=}-2221\text{kJ}/\text{mol}$

From above reaction equation 1 mole propane reacts with 5 mole oxygen and form $2221\text{kJ}/\text{mol}$ heat. Negative sign indicates heat is released in the process.

Molar mass of propane is $\text{44}\text{.1}\text{g}/\text{mol}$ .

  $\begin{array}{c}\text{Moles=}\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{5g}}{\text{44}\text{.1}\text{g}/\text{mol}}\\ \text{=0}\text{.1134mol}\end{array}$

Calculate the amount of heat release when 0.1134 mol propane reacts with excess oxygen.

  $\text{0}\text{.1134}\cancel{\text{mol}}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{×}\frac{\text{2221kJ}}{\text{1}\cancel{\text{mol}}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}=251.86\text{kJ}$

Thus amount of heat produced from 5.00g propane is 251.86 kJ.

Conclusion

Amount of heat produced from 5.00g propane is 251.86 kJ.