Equilibrium constant for the following equilibrium is given at 0oC Na 2 HPO 4 - 12H 2 O s - Na 2 HPO 4 - 7H 2 O s - 5H 2 O s K p -31-25- 10 -13The partial pressure of water is-

The correct option is D 5× 10 ^ 3 atm Solution: (D) 5 ×10^ 3 atm Na_2HPO_4.12 H_2O_( s )⇌Na_2HPO_4.7H_2O_( s ) + 5 H_2O_( g ); K_p = 3 ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

Equilibrium c...

Question

Equilibrium constant for the following equilibrium is given at $0^{o} C$
${N a}_{2} {H P O}_{4} \cdot {12 H}_{2} O (s) ⇌ {N a}_{2} {H P O}_{4} \cdot {7 H}_{2} O (s) + {5 H}_{2} O (s)$
$K_{p} = 31.25 \times 10^{- 13}$
The partial pressure of water is:

\frac{1}{5} \times 10^{- 3} a t m

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

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5 \times 10^{- 2} a t m

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

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Solution

The correct option is D $5 \times 10^{- 3} a t m$
Solution: $(D)$ $5 \times 10^{- 3} a t m$
${{N a}_{2} H P O_{4} .12 H_{2} O}_{(s)} ⇌ {{N a}_{2} H P O_{4} .7 H_{2} O}_{(s)} + 5 {H_{2} O}_{(g)}; K_{p} = 31.25 \times 10^{- 13}$
$K_{p} = {(P_{H_{2} O})}^{5}$
$\Rightarrow {(P_{H_{2} O})}^{5} = 31.25 \times 10^{- 13}$
$\Rightarrow P_{H_{2} O} = \sqrt[5]{3125 \times 10^{- 15}}$
$\Rightarrow P_{H_{2} O} = 5 \times 10^{- 3} a t m$
Hence the partial pressure of water is $5 \times 10^{- 3} a t m$ .

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Similar questions

Q. If a mixture of

N_{2}

and

H_{2}

in the ratio 1 : 3 at 50 atmosphere and

650^{o} C

is allowed to react till equilibrium is reached. Ammonia present at equilibrium was at 25 atm pressure. Calculate the equilibrium constant (

K_{P}

) for the reaction.

N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}

Q. Equilibrium constant for the following equilibrium is given at

0^{o} C

N a_{2} H P O_{4} .12 H_{2} O ⇌ N a_{2} H P O_{4} .7 H_{2} O (s) + 5 H_{2} O (g)

K_{p} = 31.25 \times 10^{- 23}

at equilibrium what will be partial pressure of water vapour:

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

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Calculate the pressure of water vapour in equilibrium at

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Q. The following equilibrium constants were determined at

1120 K

2 C O (g) ⇌ C (s) + C O_{2} (g); K_{P_{1}} = 10^{- 14} a t m^{- 1} C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g); K_{p_{2}} = 6 \times 10^{- 3} a t m^{- 1}

What is the equilibrium constant

K_{C}

for the following reaction at

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;

C (s) + C O_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g)

The equilibrium constant of the reaction

$S O_{2 (g)} + \frac{1}{2} O_{2} (g) ⇋$ SO_3_{(g)}~is~4 \times 10^{-3}atm .The equilibrium constant of the reaction 2SO_3_(g) $⇋$ 2SO_2_{(g)}+O_2(g)\) would be :

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Equilibrium constant for the following equilibrium is given at 0oCNa2HPO4⋅12H2O(s)⇌Na2HPO4⋅7H2O(s)+5H2O(s)Kp=31.25×10−13The partial pressure of water is:

The correct option is D 5×10−3atmSolution: (D) 5×10−3atmNa2HPO4.12H2O(s)⇌Na2HPO4.7H2O(s)+5H2O(g);Kp=31.25×10−13Kp=(PH2O)5⇒(PH2O)5=31.25×10−13⇒PH2O=5√3125×10−15⇒PH2O=5×10−3atmHence the partial pressure of water is 5×10−3atm.

Equilibrium constant for the following equilibrium is given at $0^{o} C$
${N a}_{2} {H P O}_{4} \cdot {12 H}_{2} O (s) ⇌ {N a}_{2} {H P O}_{4} \cdot {7 H}_{2} O (s) + {5 H}_{2} O (s)$
$K_{p} = 31.25 \times 10^{- 13}$
The partial pressure of water is: