Equilibrium constants are given (in atm) for the following reactions at TK:SrCl2.6H2O(s)⇌ SrCl2⋅2H2O(s)+4H2O(g) ; KP=5×10−12 Na2HPO4⋅12H2O(s)⇌ Na2HPO4⋅7H2O(s)+5H2O(g) ; Kp=2.43×10−13Na2SO4⋅10H2O(s)⇌ Na2SO4(s)+10H2O(g) ; Kp=1.024×10−27.The vapour pressure of water at TK is 4.56 torr. Correct statement(s) among the following is/are

A) Among the three equilibriums, in the first equilibrium PH2O is least. Hence SrCl2.2H2O is the most effective drying agent.Na2SO4.10H2O(S)⇌ Na2SO4+10H2O(S);Kp=1.02×10−27(PH2O)10=1024×10−30⇒PH2O=2×10−3 atm =2×10−3×760=1.52torrRelative humidity =1.524.56×100= 33.3%At 33.3% R.A in air no net reaction.At R.H> 33.3%, backward reaction favourable i.e Na2SO4 absorb moisture from air.At R.H<33.3%, forward reaction favourable i.e Na2SO4.10H2O looses H2O (efflorescent)

Equilibrium constants are given (in atm) for the following reactions at TK:SrCl2.6H2O(s)⇌ SrCl2⋅2H2O(s)+4H2O(g) ; KP=5×10−12 Na2HPO4⋅12H2O(s)⇌ Na2HPO4⋅7H2O(s)+5H2O(g) ; Kp=2.43×10−13Na2SO4⋅10H2O(s)⇌ Na2SO4(s)+10H2O(g) ; Kp=1.024×10−27.The vapour pressure of water at TK is 4.56 torr. Correct statement(s) among the following is/are

A) Among the three equilibriums, in the first equilibrium PH2O is least. Hence SrCl2.2H2O is the most effective drying agent.Na2SO4.10H2O(S)⇌ Na2SO4+10H2O(S);Kp=1.02×10−27(PH2O)10=1024×10−30⇒PH2O=2×10−3 atm =2×10−3×760=1.52torrRelative humidity =1.524.56×100= 33.3%At 33.3% R.A in air no net reaction.At R.H> 33.3%, backward reaction favourable i.e Na2SO4 absorb moisture from air.At R.H<33.3%, forward reaction favourable i.e Na2SO4.10H2O looses H2O (efflorescent)

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Equilibrium constants are given (in atm) for the following reactions at TK:
$S r C l_{2} .6 H_{2} O (s) ⇌_{}^{} S r C l_{2} \cdot 2 H_{2} O (s) + 4 H_{2} O (g); K_{P} = 5 \times 10^{- 12}$
$N a_{2} H P O_{4} \cdot 12 H_{2} O (s) ⇌_{}^{} N a_{2} H P O_{4} \cdot 7 H_{2} O (s) + 5 H_{2} O (g); K_{p} = 2.43 \times 10^{- 13}$
$N a_{2} S O_{4} \cdot 10 H_{2} O (s) ⇌_{}^{} N a_{2} S O_{4} (s) + 10 H_{2} O (g); K_{p} = 1.024 \times 10^{- 27} .$
The vapour pressure of water at TK is 4.56 torr. Correct statement(s) among the following is/are

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The most effective drying agent at TK from the above data is $S r C l_{2} \cdot 2 H_{2} O$

At above 33.33 % relative humidity will $N a_{2} S O_{4}$ be deliquescent when exposed to the air at TK?

At below 33.33% relative humidity will $N a_{2} S O_{4} \cdot 10 H_{2} O$ be efflorescent when exposed to air at TK

A hydrated salt can act as efflorescent at certain reduced pressure.

answer is A, B, C, D.

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Detailed Solution

A) Among the three equilibriums, in the first equilibrium $P_{H_{2} O}$ is least. Hence $S r C l_{2} .2 H_{2} O$ is the most effective drying agent.

$N a_{2} S O_{4} .10 H_{2} O (S) ⇌_{}^{} N a_{2} S O_{4} + 10 H_{2} O (S); K_{p} = 1.02 \times 10^{- 27}$

${(P_{H_{2} O})}^{10} = 1024 \times 10^{- 30} \Rightarrow P_{H_{2} O} = 2 \times 10^{- 3} a t m = 2 \times 10^{- 3} \times 760 = 1.52 t o r r$

Relative humidity $= \frac{1.52}{4.56} \times 100 = 33.3 %$

At 33.3% R.A in air no net reaction.
At R.H> 33.3%, backward reaction favourable i.e $N a_{2} S O_{4}$ absorb moisture from air.
At R.H<33.3%, forward reaction favourable i.e $N a_{2} S O_{4} .10 H_{2} O$ looses $H_{2} O$ (efflorescent)