Question

0.1 moles of ethane gas $\left(C_2{H}_6\right)$ and 0.3 moles of oxygen gas are sealed in a flask at $27{}^{o}C$ and 1.0 atm pressure. The flask is heated to 1000 K and only the following reaction takes place:
$C_2{H}_6+\frac{5}{2}{O}_2\to 2CO+3H_{2}O$

Calculate the partial pressure of left over reactants when the mixture is cooled down to $27{}^{o}C$.

• A. 0.267 atm
• B. 0.667 atm
• C. 0.038 atm
• D. 0.125 atm

Answer 1

The correct option is D 0.125 atm

1 mole of ethane needs 2.5 moles of oxygen, hence 0.1 moles will need 0.25 moles. Hence the limiting reagent is ethane. The final composition of the products at 1000 K is given by:
$C_2{H}_6=0\text{mol}{O}_2=0.05\text{mol}CO=0.20\text{mol}{H}_{2}O=0.30\text{mol}$
$\text{Total}=0.55\text{mol}$
$P_{1}V=n_{1}R{T}_1{P}_{2}V=n_{2}R{T}_2{P}_2=\frac{n_2{T}_2}{n_1{T}_1}\times P_1=\frac{0.55\times 1000}{0.4\times 300}\times 1=\frac{55}{12}\text{atm}$

When the mixture is cooled back down, the number of moles of gases would be 0.25 mol since water becomes a liquid.

Total pressure at $27{}^{o}C$:
$P_3=\frac{0.25}{0.4}\times P_1=0.625\text{atm}$
$p_{O_2}=\frac{0.05}{0.25}\times 0.625=0.125\text{atm}$