Question

Calculate the change in pressure (in atm) where 2 mole of ${\text{NO}}$ and 16g of ${{\text{O}}_{\text{2}}}$ , in a ${\text{6}}{\text{.25}}$ litre contain originally at ${{27^\circ C}}$ react to produce the maximum quantity of ${\text{NO}}$ , possible according to the equation:
${\text{2NO(g) + }}{{\text{O}}_{\text{2}}}{\text{(g) }} \to {\text{ 2N}}{{\text{O}}_{\text{2}}}{\text{(g)}}$ (Take R= $\dfrac{{\text{1}}}{{{\text{12}}}}{\text{lt atm /mol K}}$ )

Answer 1

Hint In the above question, we are asked to find out the change in pressure when initial and final conditions are given. For this, we will find individual pressure using the ideal gas equation and find the difference to get the desired result.
Formula Used
${\text{PV = nRT}}$
Where P= pressure applied, V= volume of gas present, n= total number of moles present, R= universal gas constant, T= temperature.

Complete step by step solution:
In the above question, since we are provided with moles of ${\text{NO}}$ and weight of ${{\text{O}}_{\text{2}}}$ is given, we have to first calculate the moles of oxygen present in order to check which of the two reactant acts as limiting agent.
We know that ${\text{n = }}\dfrac{{\text{m}}}{{\text{M}}}$
Where n= number of moles, m=given mass, M= molar mass.
Molar mass of oxygen = 2 $\times$ atomic mass of oxygen = ${{2 \times 16 = 32g}}$
Hence, number of moles of oxygen= $\dfrac{{\text{m}}}{{\text{M}}}{\text{ = }}\dfrac{{{\text{16}}}}{{{\text{32}}}}{\text{ = 0}}{\text{.5}}$
According to the balance chemical equation given in the question:
${\text{2NO(g) + }}{{\text{O}}_{\text{2}}}{\text{(g) }} \to {\text{ 2N}}{{\text{O}}_{\text{2}}}{\text{(g)}}$
2 moles of ${\text{NO}}$ reacts with 1 mole of ${{\text{O}}_{\text{2}}}$ .
Since, only ${\text{0}}{\text{.5 moles}}$ of ${{\text{O}}_{\text{2}}}$ is present hence, it acts as a limiting reagent.
So, $0.5$ moles of ${{\text{O}}_{\text{2}}}$ reacts with 1 mole of ${\text{NO}}$ to give 1 mole of ${\text{N}}{{\text{O}}_{\text{2}}}$ .
Initial condition:
At first, 2 moles of ${\text{NO}}$ and ${\text{0}}{\text{.5 moles}}$ of ${{\text{O}}_{\text{2}}}$ is present.
So, n = ${\text{2 + 0}}{\text{.5 = 2}}{\text{.5}}$
According to ideal gas equation:
${\text{PV = nRT}}$
By rearranging we get:
${{\text{P}}_{\text{i}}}{\text{ = }}\dfrac{{{\text{nRT}}}}{{\text{V}}}{\text{ = }}\dfrac{{{\text{2}}{\text{.5RT}}}}{{\text{V}}}$
FInal condition:
After reaction, whole oxygen get used up, so 1 mole of ${\text{NO}}$ and 1 mole of ${\text{N}}{{\text{O}}_{\text{2}}}$ is present.
So, n = ${\text{1 + 1 = 2}}$
According to ideal gas equation:
${\text{PV = nRT}}$
By rearranging we get:
${{\text{P}}_{\text{f}}}{\text{ = }}\dfrac{{{\text{nRT}}}}{{\text{V}}}{\text{ = }}\dfrac{{{\text{2RT}}}}{{\text{V}}}$
So, change in pressure ${{\Delta P = }}{{\text{P}}_{\text{i}}}{\text{ - }}{{\text{P}}_{\text{f}}}{\text{ = }}\dfrac{{{\text{2}}{\text{.5RT}}}}{{\text{V}}}{\text{ - }}\dfrac{{{\text{2RT}}}}{{\text{V}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.5RT}}}}{{\text{V}}}$
T= ${{27^\circ C = 27 + 273 = 300K}}$
V= ${\text{6}}{\text{.25L}}$
R= $\dfrac{1}{{12}} = 0.083$
Substituting the values, we get:
${{\Delta P = }}\dfrac{{{\text{0}}{\text{.5RT}}}}{{\text{V}}}{\text{ = }}\dfrac{{{\text{0}}{{.5 \times 0}}{{.083 \times 300}}}}{{{\text{6}}{\text{.5}}}}{\text{ = 1}}{\text{.9}} \simeq {\text{2}}$
Therefore, change in pressure is 2 atm.

Note:
In these types of questions where weight or moles of each reactant is given, we have first found out limiting reagents. And according to that limiting reagent concentration we have to do further calculations.