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Standard XII

Chemistry

Osmotic Pressure

A compound ...

Question

A compound $A$ dissociate by two parallel first order paths at certain temperature
$A (g) k_{1} ({m i n}^{- 1}) - ----- \to 2 B (g)$ $k_{1} = 6.93 \times 10^{- 3} m i n^{- 1}$
$A (g) k_{2} ({m i n}^{- 1}) - ----- \to C (g)$ $k_{2} = 6.93 \times 10^{- 3} m i n^{- 1}$
The reaction started with $1$ mole of pure ' $A$ ' in $1$ litre closed container with initial pressure $2$ atm. What is the pressure (in atm) developed in container after $50$ minutes from start of experiment?

1.25

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0.75

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1.50

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2.50

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Solution

The correct option is D $2.50$
For both reactions, half life is 100 min so after 50 min no. of moles of all reagents in solution is 1.25 mole. So the pressure of the container is $2 \times 1.25 / 1 = 2.50$ atm.

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The correct option is D 2.50For both reactions, half life is 100 min so after 50 min no. of moles of all reagents in solution is 1.25 mole. So the pressure of the container is 2×1.25/1=2.50 atm.

The correct option is D $2.50$
For both reactions, half life is 100 min so after 50 min no. of moles of all reagents in solution is 1.25 mole. So the pressure of the container is $2 \times 1.25 / 1 = 2.50$ atm.