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Standard XII

Chemistry

Ideal Gas Law

A mixture in ...

Question

A mixture in which the mole ratio of $H_{2}$ and $O_{2}$ is $2 : 1$ is used to prepare water by the reaction:
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} (g) O$

The total pressure in the container is $0.8$ atm at 20 $^{o} C$ before the reaction. The final pressure (in atm) at $120$ $^{o} C$ after the reaction, assuming $80 %$ yield of water is (as nearest integer) :

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Solution

The reaction is as follows:
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} (g) O$
Initial moles: $2 a$ $a$ $0$
Final moles: $(2 a - 2 x)$ $(a - x)$ $2 x$
Given, $2 x = 2 \times \frac{80}{100} \times 1.6 a$
$∴ x = 0.8 a$
Thus, after the reaction
left $H_{2} = 2 a - 1.6 a = 0.4 a$ mol
left $O_{2}$ = 0.2a mol
formed $H_{2} O = 1.6 a$ mol
Total moles after the reaction at $120^{o} C$ in gaseous phase $= 0.4 a + 0.2 a + 1.6 a = 2.2 a$
At initial conditions, P=0.8 atm , T= 293K
Total moles = $3 a$
from $P V = n R T$ , we get
$0.8 \times V = 3 a \times R \times 293$
$V = \frac{3 a \times R \times 293}{0.8}$
Now using PV=nRT after the reaction,volume remains constant
$\frac{P \times 3 a \times R \times 293}{0.8} = 2.2 a \times R \times 393$
$P = 0.787$ atm
So, the answer is $1$ atm.

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