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

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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$

$x=0.8a$

$\therefore$ 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.

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