Question

Consider the reaction:

$2X\left(g\right)+3Y\left(g\right)\to Z\left(g\right)$

Where gases $X$ and $Y$ are insoluble and inert to water and $Z$ is soluble in water, it forms a basic solution. In an experiment $3$ mole each of $X$ and $Y$ are allowed to react in $15L$ flask at $500K$. When the reaction is complete, $5L$ of water is added to the flask and temperature is reduced to $300K$. The pressure in the flask is:

[Neglect aqueous tension]

• A. 1.64 atm
• B. 2.46 atm
• C. 4.92 atm
• D. 3.28 atm

Answer 1

Answer: (B)

2.46 atm

The reaction is $2X\left(g\right)+3Y\left(g\right)\to Z\left(g\right)$.

$2$ moles of $X$ react with $3$ moles of $Y$ to form $1$ mole of $Z$.

Now $3$ moles of $X$ and $3$ moles of $Y$ are present in the flask. $Y$ is the limiting reagent.

After reaction $1$ mole of $X$ is remaining and $1$ mole of $Z$ is formed. $X$ is insoluble in water whereas $Z$ is soluble in water.

Thus the pressure in the flask is due to $1$ mole of $X$.

The total volume of a flask is $15L$. The volume of water is $5L$.

Thus the volume of gas is $15-5=10L$.

The ideal gas equation is $PV=nRT$.

Hence, $P=\frac{nRT}{V}=\frac{1\times .0821\times 300}{10}=2.46atm$.