Question

One litre of $O _{2}$ and one litre of $H _{2}$ are taken in a vessel of 2-litre capacity at S.T.P. The gases are made to combine by applying an electric spark.

Assume that water is formed quantitatively. How many grams of water are formed? What is the other component present in the vessel and in what weight?

If the vessel is now heated to $100^{\circ} C$, what will be the total pressure inside the vessel in $mm$ of $Hg$?

Answer 1

$\quad 2 H _{2}( g )+ O _{2}( g )  =2 H _{2} O ( l ) \\2 \times 22 \cdot 4 \text { litres } \quad 22 \cdot 4 \text { litres }  2 \times 18 g =36 g$

It means that $1 \cdot 0$ litre $H _{2}$ and $0.5$ litre $O _{2}$ react to form water.

$\because 2 \times 22.4$ litres $H_{2}$ at S.T.P. produce $36 g$ water

$\therefore \quad 1 \cdot 0$ litre $H_{2}$ at S.T.P. will give $\frac{36}{2 \times 22 \cdot 4} g = 0 \cdot 8 0 3 5 g$ water.

Vol. of $O _{2}$ left unreacted $=(1 \cdot 0-0 \cdot 5)$ litre $=0 \cdot 5$ litre at STP.

$\because 22 \cdot 4$ litres $O_{2}$ at S.T.P. weigh $32 g$

$\therefore 0.5$ litre $O_{2}$ at S.T.P. will weigh $\frac{32 \times 0.5}{22.4} g = 0 \cdot 7142 g$.

The volume of the vessel is 2 litres. At $100^{\circ} C$, liquid water will become vapour and it will also exert pressure. Thus the vessel contains water vapour and $O_{2}$ at $100^{\circ} C$.

We know from gas equation that $P V=n R T$.

$R=0 \cdot 082$ litre-atm $K ^{-1} mol ^{-1} \quad=\frac{1}{22 \cdot 4}+\frac{1}{44 \cdot 8}=\frac{3}{44 \cdot 8}=n$

$T=(100+273) K =373 K ; \quad \because P V=n R T$

$\therefore \quad P \times 2=\frac{3}{44 \cdot 8} \times 0.082 \times 373 atm$

$\therefore \quad P=\frac{3 \times 0.082 \times 373}{2 \times 44.8} atm =\frac{3 \times 0.082 \times 373 \times 760}{2 \times 44.8} mm$ of $Hg .$

$=778 \cdot 3 mm$ of $Hg$.