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Chemistry - Chemical Equilibrium Question with Solution | TestHub

ChemistryChemical EquilibriumMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

In reaction $N_{2} O_{4} (g) \to 2 N O_{2} (g)$ , The observed molecular weight $80 g m o l^{- 1}$ at 350 K. The percentage dissociation of $N_{2} O_{4} (g)$ at 350 K is

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Solution:

Or $\begin{matrix} N_{2} O_{4} \to & 2 N O_{2} \\ 1 & 0 \\ 1 - α & 2 α \end{matrix})$
Mass remains same
Average molar man $= \frac{M_{N_{2} O_{4}}}{(1 + α)}$
So, $= \frac{28 + 64}{(1 + α)} = \frac{92}{(1 + α)}$
$1 + α = \frac{92}{80} = \frac{23}{20}$
$α = \frac{23}{20} - 1 = \frac{3}{20}$
$% α = \frac{3}{20} \times 100 = 15 %$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Chemical EquilibriumSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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