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Reaction Quotient

For the equil...

Question

For the equilibrium, $L i C l \cdot 3 N H_{3} (s) ⇌ L i C l \cdot N H_{3} (s) + 2 N H_{3} (g)$ ; $K_{p} = 9$ $a t m^{2}$ at $37^{o}$ C. A $5$ litre vessel contains $0.1$ mole of $L i C l \cdot N H_{3}$ . How many moles of $N H_{3}$ should be added to the flask at this temperature to derive the backward reaction for completion?
Use: $R = 0.082$ atm-L/mol K

0.2

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0.59

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0.69

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0.79

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Solution

The correct option is A $0.2$

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Similar questions

L i C l \cdot {3 N H}_{3} (s) ⇌ L i C l \cdot {N H}_{3} (s) + 2 {N H}_{3} (g)

K_{p} = 9 {a t m}^{2}

37^{o} C

5

0.1

L i C l \cdot {3 N H}_{3}

{N H}_{3}

R = 0.082 a t m . L / m o l K

L i C l . 3 N H_{3} (s) ⇌ L i C l . N H_{3} (s) + 2 N H_{3} (g);

K_{p} = 9 a t m^{2}

37^{\circ} C .

5 L

0.1

L i C l . N H_{3}

N H_{3}

R = 0.082 {atm L mol}^{- 1} K^{- 1}

L i C l .3 {N H}_{3} (s) ⇌ L i C l .3 {N H}_{3} (s) + 2 {N H}_{3}

K_{p}

{a t m}^{2}

40 C

L i C l .3 N H_{3}

N H_{3}

L i C l . 3 N H_{3} (s) ⇌ L i C l . N H_{3} (s) + 2 N H_{3} (g);

K_{p} = 9 a t m^{2}

37^{\circ} C .

5 L

0.1

L i C l . N H_{3}

N H_{3}

R = 0.082 a t m L / m o l K

27^{o} C .

L i C l .3 N H_{3} (s) ⇌ L i C l . N H_{3} (s) + 2 N H_{3} (g); K_{P} = 9 a t m^{2}

L i C l . N H_{3} .

N H_{3}

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