Question

1 mol of $O_2$ and 3 mol of $H_2$ are mixed isothermally in a $125.3\text{L}$ container at $125{}^{\circ }\text{C}$. The gaseous mixture undergoes reaction to form water vapour. Assuming ideal gas behaviour, the total pressure before and after the formation of $H_{2}O\left(g\right)$ respectively is:

• A. 1.056 bar, 1.056 bar
• B. 1.056 bar, 0.792 bar
• C. 0.792 bar, 1.3056 bar
• D. 0.264 bar, 0.528 bar

Answer 1

The correct option is B 1.056 bar, 0.792 bar

Initially total moles of the gaseous mixture $(O_2$ and $H_2)=4$ mol

Thus, $P_{total}=\frac{nRT}{V}=\frac{4\times 0.0831\times 398}{125.3}=1.056\text{bar}$

$\begin{array}{cccccc}& 2H_2\left(g\right)& +& O_2\left(g\right)& \to & 2H_{2}O\left(g\right)\\ \text{Initial moles}& 3& & 1& & 0\\ \text{Reacted/formed moles}& 2& & 1& & 2\\ \text{remaining moles}& 1& & 0& & 2\end{array}$

Thus, the total moles after reaction $=n_{H_2}+n_{H_{2}O}=1+2=3$ mol

$P_{total}=\frac{nRT}{V}=\frac{3\times \mathrm{0.0831.4}\times 398}{125.3}=0.792\text{bar}$