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Question: \(20%\text{ of }{{\text{N}}_{2}}{{O}_{4}}\) molecules are dissolved in a sample of gas at\(27{}^\cir...

2020%\text{ of }{{\text{N}}_{2}}{{O}_{4}} molecules are dissolved in a sample of gas at27∘C and 760torr⁡27{}^\circ C\text{ and }760\operatorname{torr}. Mixture has the density at equilibrium equal to:
A. 1.48g⁡/L1.48\operatorname{g}/L
B. 1.84g⁡/L1.84\operatorname{g}/L
C. 2.25g⁡/L2.25\operatorname{g}/L
D. 3.12g⁡/L3.12\operatorname{g}/L

Explanation

Solution

The density of the mixture includes gases on both sides of the equilibrium. Use the ideal gas law which has density as one of the variables.

Complete step by step answer:
The ideal gas law as we know it is: PV=nRTPV=nRT, where
- “P” is pressure in atm⁡\operatorname{atm}
- “V” is volume in litreslitres
- “n” is the number of moles
- “R” and “T” are the universal gas constant and temperature in kelvin respectively
The ideal gas equation which has density as one of its variable is: PM=dRTPM=dRT
Where “M” and “d” are molar mass and the density of the gaseous reactants present, respectively.
To use the above equation we have to find the total molar mass of the reactants and products present at equilibrium. It goes as follows:
The reaction is given as-
N2O4(g)⇌2NO2(g){{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)

Let the initial concentration of N2O4{{N}_{2}}{{O}_{4}} be α\alpha , that will make the initial concentration of NO2N{{O}_{2}} to be zero as the reaction has not started yet. It is given that at equilibrium 20% ofN2O4{{N}_{2}}{{O}_{4}} is dissociated, which means that the concentration of the reactant will decrease by 20% and that of the product will increase by 20%. Twenty percent of “α\alpha ” is 0.2α0.2\alpha ; so the reactant becomes 0.8α0.8\alpha and the product becomes 0.4α0.4\alpha (as the product has a higher stoichiometric coefficient). The overall concentration along with the equilibrium is given as follows:

& \text{ }{{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g) \\\ & Initial:\text{ }\alpha \text{ }0 \\\ & At\text{ equilibrium}:\text{ }0.8\alpha \text{ }0.4\alpha \\\ \end{aligned}$$ That makes the total number of moles at equilibrium as$1.2\alpha $ (adding $0.8\alpha +0.4\alpha =1.2\alpha $) Using mole fraction we can find the total mass of the mixture at equilibrium. It is as follows: $$\begin{aligned} & \operatorname{Total}\operatorname{Mass}(M)={{\chi }_{{{N}_{2}}{{O}_{4}}}}\times {{M}_{{{N}_{2}}{{O}_{4}}}}+{{\chi }_{N{{O}_{2}}}}+{{M}_{N{{O}_{2}}}} \\\ & \Rightarrow M=\dfrac{0.8\alpha }{1.2\alpha }\times 92+\dfrac{0.4\alpha }{1.2\alpha }\times 46 \\\ & \Rightarrow M=60.72+15.18 \\\ & \Rightarrow M=75.9g \\\ \end{aligned}$$ Applying the equation for ideal gas law- $d=\dfrac{PM}{RT}$ Entering the respective values as we have found out above, the above equation becomes: $$d=\dfrac{1\times 75.9}{0.0821\times 300}$$ Because $P=760\operatorname{torr}=1\operatorname{atm}$ And $$R=0.0821\dfrac{\operatorname{atm}L}{molK}$$ , $T=27{}^\circ C=300\operatorname{K}$ So we get $d=3.081\operatorname{g}/L$which is nearly equal to the option (D)$3.12\operatorname{g}/L$. **So, the correct answer is “Option D”.** **Note:** Taking care of the units matters a lot in this type of questions. Such as in this case the pressure is given in “$\operatorname{torr}$” but in the formula you have to apply by converting it into “$\operatorname{atm}$”. Same is true for temperature. The universal gas constant has two values which are commonly used depending on the units given in the question. They are as follows: \- $R=0.0821\dfrac{L\operatorname{atm}}{mol\operatorname{K}}$ \- $R=8.314\dfrac{J}{\operatorname{mol}K}$ Care should be taken while applying the correct value in the solution.