# A mixture in which the mole ratio of $H_2$ and $O_2$ is 2 : 1 is used to prepare water by the reaction $2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$.The total pressure in the container is 0.8atm at $20^{\large\circ}C$ before the reaction.Determine the final pressure at $120^{\large\circ}C$ after reaction assuming 80% yield of water.

$(a)\;2.3atm\qquad(b)\;0.787atm\qquad(c)\;1.234atm\qquad(d)\;0.024atm$

Step 1:
$2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$
Initial mole :
$2H_2(g)\Rightarrow 2a$
$O_2(g)\Rightarrow a$
$2H_2O(g)\Rightarrow 0$
Final mole :
$2H_2(g)\Rightarrow 2a-2x$
$O_2(g)\Rightarrow a-x$
$2H_2O(g)\Rightarrow 2x$
Step 2:
$2x=2a\times \large\frac{80}{100}$$=1.6a x=0.8a \therefore after the reaction H_2 left =2a-1.6a \Rightarrow 0.4a mole O_2 left =0.2a mole H_2O formed =1.6a mole Step 3: Total mole at 120^{\large\circ}C in gaseous phase =0.4a+0.2a+1.6a \Rightarrow 2.2a Given at initial conditions P=0.8atm,T=293K P\times V=nRT 0.8\times V=3a\times R\times 293 V=\large\frac{3a\times R\times 293}{0.8} Step 4: The volume of container remains constant P\times V=nRT P\times\large\frac{3a\times R\times 293}{0.8}$$=2.2a\times R\times 393$
$P=\large\frac{393\times 0.8\times 2.2}{3\times 293}$
$\Rightarrow 0.787atm$
Hence (b) is the correct answer.