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:
$2{H_2}(g) + {O_2}(g) \to 2{H_2}O(g)$
The total pressure in the container is $0.8atm$ at ${20^0}C$ before the reaction. The final pressure (in atm) at ${120^0}C$ after the reaction, assuming $80\% $ yield of water is (as nearest integer).

Answer 1

Hint: In order to deal with this question, we will consider the moles of each component before the reaction and after the reaction, we will apply the ideal gas equation after receiving total moles which will give us the required result.

Complete step by step solution:

The reaction consists of:

                            $2{H_2}(g) + {O_2}(g) \to 2{H_2}O(g)$
Initial mole: $\;\;\;\;\;\;2a \;\;\;\;\;\;\;\;\;\;\;\; a \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0$
Final mole: $\;\;\;\;2a - 2x \;\;\;\;\; a - x \;\;\;\;\;\;\;\;\;\;\,\; 2x$

Given, $2x = 2a \times 80 \div 100 = 1.6a$
So, $x = 0.8a$
After the reaction ${H_2}$ left $ = 2a - 1.6a = 0.4a$mole
And ${O_2}$ left $ = 0.2a$ mole
${H_2}O$ formed $ = 1.6a$ mole
Total moles after the reaction at ${120^0}C$ in gaseous phase $ = 0.4a + 0.2a + 1.6a = 2.2a$

As we know the ideal equation for gas is given as
$PV = nRT$
In the beginning, $P = 0.8$ atm , $T = 293K$
Overall Moles = 3a
So that we can write
$
  0.8 \times V = 3a \times R \times 293 \\
  V = (3a \times R \times 293) \div 0.8 \\
 $
Because the volume of the container will not change.
So by using the gas equation
$P \times (3a \times R \times 293) \div 0.8 = 2.2a \times R \times 393$
$P = (393 \times 0.8 \times 2.2) \div (3 \times 293) = 0.787$atm
Therefore, the pressure required is 0.787atm.

Additional Information-
The law of multiple proportion states that when two components combine to form more than one substance, the mass of one part, together with the fixed mass of the other element, will always be the ratio of the whole quantity.
The law of definite proportion- This law states that for a given compound, the chemical compound always contains its component elements in the fixed ratio by mass and does not depend on its source or method of preparation.

Note: The mole is the unit of measurement for the amount of the substance in the International System of Units(SI) 1 mole contains $6.023 \times {10^{23}}$ particles. With the help of the gas equation we can find the volume covered by gas particles and the implied pressure of those particles. Here R stands for the universal gas constant which is equal to $8.314J{K^{ - 1}}{(mol)^{ - 1}}$.