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Law of Mass Action

Three moles o...

Question

Three moles of $N_{2}$ react with $2$ of $O_{2}$ in a two-liter container to form one mole of NO. What will be the number of moles of $O_{2}$ per litre at equilibrium?

0.75

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1.5

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1.25

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None of above

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Solution

The correct option is A 0.75
$N_{2} + O_{2} \to 2 N O$
Given
Initial moles of $N_{2} = 3$
Initial moles of $O_{2} = 2$
Volume of container = 2L
Moles of $N O$ formed at equillibrium = 1

From the balanced reaction we can say that 1mol $O_{2}$ reacts to form 2 mol of $N O$ .
Therefore by unitary method number of moles of $O_{2}$ required to form 1 mol $N O = 0.5$ .
Therefore number of moles of $O_{2}$ remaining at equillibrium $= 2 - 0.5 = 1.5 m o l e s .$
Therefore molar concentration of $O_{2}$ = $\frac{moles of O_{2}}{Volume of concentration} = 1.5 / 2 = 0.75 m o l / L .$

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Three moles of N2 react with 2 of O2 in a two-liter container to form one mole of NO. What will be the number of moles of O2 per litre at equilibrium?

Three moles of $N_{2}$ react with $2$ of $O_{2}$ in a two-liter container to form one mole of NO. What will be the number of moles of $O_{2}$ per litre at equilibrium?