Question

Consider the following questions:

(a) What is the total volume of the CO₂(g) and H₂O(g) at 600 C and 0.888 atm produced by the combustion of 1.00 L of C₂H₆ measured at STP?

(b) What is the partial pressure of H₂O in the product gases?

Short answer

(a) The total volume of the $$\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$$ and $${\mathrm{H}}_2\mathrm{O}(\mathrm{g})$$ is $$\mathrm{V}=18\mathrm{L}$$.

(b) The partial pressure of $${\mathrm{H}}_2\mathrm{O}(\mathrm{g})$$ in the product gases is $$\mathrm{P}=0.533\mathrm{a}\mathrm{t}\mathrm{m}$$.

Step-by-step solution

Concept Introduction

Each gas exerts pressure in a container filled with more than one gas, which is known as partial pressure. The partial pressure of any gas within the container is its pressure.

Finding the Total Volume

(a)

The balanced equation is as follows –

$${\mathrm{C}}_2{\mathrm{H}}_6(\mathrm{g})+\frac{7}{2}{\mathrm{O}}_2(\mathrm{g})\to 2\mathrm{C}{\mathrm{O}}_2(\mathrm{g})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$$$$1\mathrm{m}\mathrm{o}\mathrm{l}+3.5\mathrm{m}\mathrm{o}\mathrm{l}-2\mathrm{m}\mathrm{o}\mathrm{l}+3\mathrm{m}\mathrm{o}\mathrm{l}$$

First, calculate the number of moles of$${\mathrm{C}}_2{\mathrm{H}}_6$$–

At STP,$$\mathrm{P}=1\mathrm{a}\mathrm{t}\mathrm{m},\mathrm{T}=273\mathrm{K},\mathrm{R}=0.08206\mathrm{L}\cdot \mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K},\mathrm{V}=1\mathrm{L}$$.

Number of moles of$${\mathrm{C}}_2{\mathrm{H}}_6$$–

$$\begin{array}{r}{\mathrm{n}}_{\left({\mathrm{C}}_2{\mathrm{H}}_6\right)}=\frac{\mathrm{P}\mathrm{V}}{\mathrm{R}\mathrm{T}}\\ =\frac{1.00\mathrm{a}\mathrm{t}\mathrm{m}\times 1.00\mathrm{L}}{0.08206\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}\times 273\mathrm{K}}\\ =0.0446\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$$

From the equation, it can be seen that one mole of$${\mathrm{C}}_2{\mathrm{H}}_6$$will produce five moles of$$\mathrm{C}{\mathrm{O}}_2$$and$${\mathrm{H}}_2\mathrm{O}$$. Number of moles of$$\mathrm{C}{\mathrm{O}}_2$$and$${\mathrm{H}}_2\mathrm{O}$$produced from$$0.0446\mathrm{m}\mathrm{o}\mathrm{l}$$$${\mathrm{C}}_2{\mathrm{H}}_6$$–$$0.0446\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{C}}_2{\mathrm{H}}_6\times \frac{5\mathrm{m}\mathrm{o}\mathrm{l}\left(\mathrm{C}{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}\right)}{1\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{C}}_2{\mathrm{H}}_6}=0.223\mathrm{m}\mathrm{o}\mathrm{l}\left(\mathrm{C}{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}\right)$$$$\mathrm{T}=600^{{}^{\circ }}\mathrm{C}+273=873\mathrm{K},\mathrm{P}=0.888\mathrm{a}\mathrm{t}\mathrm{m}$$

Finally, use the ideal gas law to find volume of$$\mathrm{C}{\mathrm{O}}_2$$and$${\mathrm{H}}_2\mathrm{O}$$–

$$\begin{array}{r}\mathrm{V}=\left(\frac{\mathrm{n}\mathrm{R}\mathrm{T}}{\mathrm{P}}\right)\\ =\frac{0.223\mathrm{m}\mathrm{o}\mathrm{l}\times 0.08206\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}\times 873\mathrm{K}}{0.888\mathrm{a}\mathrm{t}\mathrm{m}}\\ =18\mathrm{L}\end{array}$$

Therefore, the value for volume is obtained as $$\mathrm{V}=18\mathrm{L}$$.

Partial Pressure of Water

(b)

From the equation, it can be seen that one mole of $${\mathrm{C}}_2{\mathrm{H}}_6$$ will produce three moles of $${\mathrm{H}}_2\mathrm{O}$$. Number of moles of $${\mathrm{H}}_2\mathrm{O}$$ produced from $$0.0446\mathrm{m}\mathrm{o}\mathrm{l}$$ $${\mathrm{C}}_2{\mathrm{H}}_6$$–

$$0.0446\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{C}}_2{\mathrm{H}}_6\times \frac{3\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{H}}_2\mathrm{O}}{1\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{C}}_2{\mathrm{H}}_6}=0.134\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{H}}_2\mathrm{O}$$

$$\mathrm{T}=600^{{}^{\circ }}\mathrm{C}+273=873\mathrm{K},\mathrm{P}=0.888\mathrm{a}\mathrm{t}\mathrm{m},\mathrm{V}=18\mathrm{L}$$

Finally, use the ideal gas law to find partial pressure of$${\mathrm{H}}_2\mathrm{O}$$–

$$\begin{array}{r}\mathrm{P}=\left(\frac{\mathrm{n}\mathrm{R}\mathrm{T}}{\mathrm{V}}\right)\\ =\frac{0.134\mathrm{m}\mathrm{o}\mathrm{l}\times 0.08206\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}\times 873\mathrm{K}}{18\mathrm{L}}\\ =0.533\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$$

Therefore, the value for partial pressure is obtained as$$\mathrm{P}=0.533\mathrm{a}\mathrm{t}\mathrm{m}$$.