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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.


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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}$$=1.6a$
$\therefore$ after the reaction $H_2$ left =$2a-1.6a$
$\Rightarrow 0.4a$ mole
$O_2$ left =0.2a mole
$H_2O$ formed =1.6a mole
Step 3:
Total mole at $120^{\large\circ}C$ in gaseous phase =$0.4a+0.2a+1.6a$
$\Rightarrow 2.2a$
Given at initial conditions
$P\times V=nRT$
$0.8\times V=3a\times R\times 293$
$V=\large\frac{3a\times R\times 293}{0.8}$
Step 4:
The volume of container remains constant
$P\times V=nRT$
$P\times\large\frac{3a\times R\times 293}{0.8}$$=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.
answered Nov 6, 2013 by sreemathi.v

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