Equilibrium constants are given in atm for the following reactions at 0- C SrCl 2 -6 H 2 O s - SrCl 2 -2 H 2 O s-4 H 2 O g -K p -6-89 - 10-12The vapour pressure of water at 0- C is 4-58 torr- Calculate the pressure of water vapour in equilibrium at 0- CA- 2-6 - 10-3 atmB- 1-62 - 10-3 atmC- 1-20 - 10-4 atmD- 4-28 - 10-3 atm

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Standard XII

Chemistry

Derivation of Kp and Kc

Equilibrium c...

Question

Equilibrium constants are given (in atm) for the following reactions at $0^{o} C$ $S r C l_{2} . 6 H_{2} O_{(s)} ⇌ S r C l_{2} . 2 H_{2} O_{(s)} + 4 H_{2} O_{(g)},$
$K_{p} = 6.89 \times 10^{- 12}$
The vapour pressure of water at $0^{o} C$ is 4.58 torr.
Calculate the pressure of water vapour in equilibrium at $0^{o} C$

2.6 \times 10^{- 3} a t m

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1.62 \times 10^{- 3} a t m

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1.20 \times 10^{- 4} a t m

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4.28 \times 10^{- 3} a t m

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Solution

The correct option is B $1.62 \times 10^{- 3} a t m$
K_p is equilibrium constant used when equilibrium concentrations are expressed in atmospheric pressure. Solid components doesnot contribute any pressure to the system. So,
$K_{P} = P_{H_{2} O}^{4} = 6.89 \times 10^{- 12}$
$∴ P_{H_{2} O} = 1.62 \times 10^{- 3} a t m$

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Similar questions

Q. Equilibrium constants are given (in atm) for the following reactions at

0^{o} C

S r C l_{2} . 6 H_{2} O_{(s)} ⇌ S r C l_{2} . 2 H_{2} O_{(s)} + 4 H_{2} O_{(g)},

K_{p} = 6.89 \times 10^{- 12}

The vapour pressure of water at

0^{o} C

is 4.58 torr.
Calculate the pressure of water vapour in equilibrium at

0^{o} C

Q. For the equilibrium:

S r C l_{2} .6 H_{2} O (s) ⇌ S r C l_{2} .2 H_{2} O (s) + 4 H_{2} O (g)

the equilibrium constant

K_{p}

= 16 x 10

^{- 12}

atm

^{4}

1^{o} C

. If one litre of air saturated with water vapour at

1^{o} C

is exposed to a large quantity of

S r C l_{2} .2 H_{2} O (s)

, what weight of water vapour will be absorbed? Saturated vapour pressure of water at

1^{o} C

a 7.6 torr.

Q. For the equilibrium

S r C l_{2} .6 H_{2} O (s) ⇌ S r C l_{2} .2 H_{2} O (s) + 4 H_{2} O (g)

the equilibrium constant

K_{p} = 16 \times 10^{- 12} a t m^{4}

1^{0} C

, what weight of water vapour will be absorbed ? Saturated vapour pressure of water at

1^{0} C = 7.6 t o r

Q. Equilibrium constants is given (in atm) for the following reaction

0^{\circ} C

N a_{2} H P O_{4} . 12 H_{2} O (s) ⇌ N a_{2} H P O_{4} . 7 H_{2} O (s) + 5 H_{2} O (g); K_{p} = 2.43 \times 10^{- 13}

The vapour pressure of water at

0^{\circ} C

4.56 t o r r

.
At what relative humidities will

N a_{2} H P O_{4} .12 H_{2} O (s)

be efflorescent when exposed to air at

0^{\circ} C

Q. For the equilibrium

S r C l_{2} \cdot 6 H_{2} O (s) ⇌ S r C l_{2} \cdot 2 H_{2} O (s) + 4 H_{2} O (g)

the equilibrium constant

K_{p} = 16 \times 10^{- 12}

a t m^{4}

1^{0}

If one litre of air saturated with water vapour at

1^{0}

C is exposed to a large quantity of

S r C l_{2} \cdot 2 H_{2} O (s)

, what weight of water vapour will be absorbed?

(Saturated vapour pressure of water at

1^{0}

C=7.6 torr.)

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Derivation of Kp and Kc

Standard XII Chemistry

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Equilibrium constants are given (in atm) for the following reactions at 0oC SrCl2.6H2O(s)⇌SrCl2.2H2O(s)+4H2O(g), Kp=6.89×10−12 The vapour pressure of water at 0oC is 4.58 torr. Calculate the pressure of water vapour in equilibrium at 0oC

The correct option is B 1.62×10−3 atmKp is equilibrium constant used when equilibrium concentrations are expressed in atmospheric pressure. Solid components doesnot contribute any pressure to the system. So, KP=P4H2O=6.89×10−12 ∴PH2O=1.62×10−3atm