Chapter 8, Problem 72QAP

On complete combustion at constant pressure, a 1.00-L sample of a gaseous mixture at 0°C and 1.00 atm (STP) evolves 75.65 kJ of heat. If the gas is a mixture of ethane (C₂H₆) and propane (C₃H₈), what is the mole fraction of ethane in the mixture?

Interpretation Introduction

Interpretation:

The mole fraction of ethane in a mixture of ethane and propane gas needs to be calculated under the given conditions.

Concept Introduction:

• A thermochemical equation is a balanced stoichiometric equation which depicts the change in the reaction enthalpy i.e. the magnitude and sign of ΔH is added to the right of the equation.
• The standard enthalpy change for a reaction ${\text{ΔH}}^{\text{0}}$ is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

i.e. $\Delta {\text{H}}^0\text{ = }{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}}\text{(products) - }{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}}\text{(reactants) ------(1)}$

where n_p and n_r are the number of moles of the products and reactants

  ${\text{The values of ΔH}}_{\text{f}}^{\text{0}}\text{ are measured under standard conditions of 25}\text{C}\text{ and 1 atm pressure}$

Since ΔH° is independent of pressure and varies very slightly with temperature, we can interpret that ΔH° is equal to the reaction enthalpy,ΔH.

Answer to Problem 72QAP

Mole fraction of ethane in the gas mixture is 0.7934.

Explanation of Solution

Given:

Volume of gas mixture, V = 1.0 L

Pressure, P = 1 atm

Temperature, T = 0°C = 273 K

Heat released during combustion of the ethane + propane mixture = 75.65 J

Calculation:

Step 1: Calculate the enthalpy of combustion of one mole of ethane

The chemical reaction corresponding to the combustion of ethane is given as:

  ${\text{C}}_2{\text{H}}_6{\text{(g) + 7/2O}}_{\text{2}}\text{(g)}\to {\text{ 2CO}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{O(l) }\Delta \text{H = ? kJ}$

The amount of heat absorbed or evolved can be given in terms of the enthalpy change for the above reaction. Based on equation (1) we have:

  $\begin{array}{l}\Delta {\text{H}}^0\text{ = }{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}}\text{(products) - }{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}}\text{(reactants) }\\ {\text{= [2ΔH}}_{\text{f}}^{\text{0}}{\text{(CO}}_{\text{2}}{\text{(g)) + 3ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{O(l))] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(C}}_2{\text{H}}_6{\text{(g)) + 7/2ΔH}}_{\text{f}}^{\text{0}}{\text{(O}}_{\text{2}}\text{(g))]}\\ {\text{Based on the ΔH}}_{\text{f}}^{\text{0}}\text{ values:}\\ \Delta \text{H = [2 moles }\times (\text{-393}\text{.5 kJ/mol) + 3 moles }\times \text{(-285}\text{.8 kJ/mol)]}\\ \text{ - [1 mole }\times \text{(-}84.7\text{ kJ/mol) + 7/2 mole }\times \text{(}0\text{ kJ/mol)]}\\ \text{ = -787 kJ -857}\text{.4 kJ + 84}\text{.7 + 0 = -1560 kJ }\end{array}$

Step 2: Calculate the enthalpy of combustion of one mole of propane

The chemical reaction corresponding to the combustion of propane is given as:

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{(g) + 5O}}_{\text{2}}\text{(g)}\to {\text{ 3CO}}_{\text{2}}{\text{(g) + 4H}}_{\text{2}}\text{O(l) }\Delta \text{H = ? kJ}$

The amount of heat absorbed or evolved can be given in terms of the enthalpy change for the above reaction. Based on equation (1) we have:

  $\begin{array}{l}\Delta {\text{H}}^0\text{ = }{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}}\text{(products) - }{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}}\text{(reactants) }\\ {\text{= [3ΔH}}_{\text{f}}^{\text{0}}{\text{(CO}}_{\text{2}}{\text{(g)) + 4ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{O(l))] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{(g))+5ΔH}}_{\text{f}}^{\text{0}}{\text{(O}}_{\text{2}}\text{(g))]}\\ {\text{Based on the ΔH}}_{\text{f}}^{\text{0}}\text{ values:}\\ \Delta \text{H = [3 mole }\times (\text{-393}\text{.5 kJ/mol) + 4 mole }\times \text{(-285}\text{.8 kJ/mol)]}\\ \text{ - [1 mole }\times \text{(-}103.8\text{ kJ/mol) + 5 mole }\times \text{(}0\text{ kJ/mol)]}\\ \text{ = -1180}\text{.5 kJ -1143}\text{.2 kJ + 103}\text{.8 + 0 = -2220 kJ }\end{array}$

Step 3: Calculate the total moles of C₂H₆+C₃H₈ in the mixture

Based on the ideal gas equation:

  $\begin{array}{l}\text{PV = nRT}\\ \text{n = }\frac{\text{PV}}{\text{RT}}=\frac{\text{1}\text{.00 atm×1}\text{.0L}}{\text{0}\text{.0821L}\text{.atm/mol-K×273K}}=0.0446\text{ moles}\\ \text{i}\text{.e}{\text{. n}}_{\text{total}}{\text{ = n}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}{\text{ + n}}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}=0.0446\text{ -----(2)}\end{array}$

Step 4: Calculate the moles of C₂H₆ and C₃H₈

Let x be the number of moles of C₂H₆

Therefore, 0.0446 − x = moles of C₃H₈

From Step 1:

1 mole of C₂H₆upon combustion releases 1560 kJ of heat

Therefore, heat released by ‘x’ moles of C₂H₆ = 1560x kJ

From Step 1:

1 mole of C₃H₈upon combustion releases 2220 kJ of heat

Therefore, heat released by ‘0.0446-x’ moles of C₂H₆ = 2220(0.0446-x) kJ

It is given that the total heat released is 75.65 kJ

  $\begin{array}{l}\begin{array}{l}\text{i}\text{.e}\text{. 1560x + 99}\text{.01 – 2220x = 75}\text{.65}\hfill \\ \text{660 x = 23}\text{.36}\hfill \\ \text{x = 0}{\text{.03539 = moles of C}}_{\text{2}}{\text{H}}_{\text{6}}\hfill \end{array}\\ \text{0}\text{.0446-x = 0}\text{.0446 - 0}\text{.03539 = 0}{\text{.00921 = moles of C}}_{\text{3}}{\text{H}}_{\text{8}}\end{array}$

Step 4: Calculate the mole fraction of C₂H₆ and C₃H₈ $\begin{array}{l}{\text{Mole fraction of C}}_{\text{2}}{\text{H}}_{\text{6}}\text{ = }\frac{{\text{Moles of C}}_{\text{2}}{\text{H}}_{\text{6}}}{\text{Total moles}}=\frac{0.03539}{0.0446}=0.7934\\ {\text{Mole fraction of C}}_{\text{3}}{\text{H}}_{\text{8}}\text{ = }\frac{{\text{Moles of C}}_{\text{3}}{\text{H}}_{\text{8}}}{\text{Total moles}}=\frac{0.00921}{0.0446}=0.2065\end{array}$

Conclusion

Mole fraction of ethane in the gas mixture is 0.7934.