Chemistry - Chemical Equilibrium Question with Solution | TestHub

ChemistryChemical EquilibriumMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasynumerical

For the reaction $3 A ⇌ C + 2 D$ , initially A was taken. At equilibrium the concentration of D is twice that of A. The value of $K_{c}$ is

Answer:

4.00

Solution:

Now, $\frac{2 x}{3} = 2 \times (C - x)$
$x = 3 C - 3 x$
$C = \frac{4}{3} x$
$K_{c} = \frac{[C] {[D]}^{2}}{{[A]}^{3}}$
$K_{c} = \frac{\frac{x}{3} \times {(\frac{2 x}{3})}^{2}}{{(\frac{4}{3} x - x)}^{3}} = \frac{(\frac{4}{27})}{(\frac{1}{27})} = 4$

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

⏱ 2mℹ️ Source: PYQ_2020

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