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The liquid disilicon hexachloride reacts with water to form solid silicon dioxide, hydrogen chloride gas and hydrogen gas are
![Si{2}Cl{6} (l) + 4H{2}O (l) ------ > 2SiO{2} (s) + 6HCl (g) + H{2} (g)](https://tex.z-dn.net/?f=Si%7B2%7DCl%7B6%7D%20%28l%29%20%20%2B%20%204H%7B2%7DO%20%28l%29%20%20------%20%3E%202SiO%7B2%7D%20%28s%29%20%20%2B%20%20%20%206HCl%20%28g%29%20%20%20%20%2B%20%20%20%20%20%20H%7B2%7D%20%28g%29)
When liquid disilicon hexachloride reacts with water to form solid silicon dioxide , hydrogen chloride gas and hydrogen gas , the following chemical equation is:
![Si{2}Cl{6} (l) + H{2}O (l) ------ > SiO{2} (s) + HCl (g) + H{2} (g)](https://tex.z-dn.net/?f=Si%7B2%7DCl%7B6%7D%20%28l%29%20%20%2B%20%20H%7B2%7DO%20%28l%29%20%20------%20%3E%20SiO%7B2%7D%20%28s%29%20%20%2B%20%20%20%20HCl%20%28g%29%20%20%20%20%2B%20%20%20%20%20%20H%7B2%7D%20%28g%29)
now , we have to balance the eqaution,
Reactants product
Si 2 1
Cl 6 1
H 2 3
O 1 2
Multiply 4 with
2 with
and 6 with HCL, we get
![Si{2}Cl{6} (l) + 4H{2}O (l) ------ > 2SiO{2} (s) + 6HCl (g) + H{2} (g)](https://tex.z-dn.net/?f=Si%7B2%7DCl%7B6%7D%20%28l%29%20%20%2B%20%204H%7B2%7DO%20%28l%29%20%20------%20%3E%202SiO%7B2%7D%20%28s%29%20%20%2B%20%20%20%206HCl%20%28g%29%20%20%20%20%2B%20%20%20%20%20%20H%7B2%7D%20%28g%29)
now ,
Reactant Product
Si 2 2
Cl 6 6
H 8 8
O 4 4
Thus, we get the desire equtions.
The liquid disilicon hexachloride reacts with water to form solid silicon dioxide, hydrogen chloride gas and hydrogen gas are
![Si{2}Cl{6} (l) + 4H{2}O (l) ------ > 2SiO{2} (s) + 6HCl (g) + H{2} (g)](https://tex.z-dn.net/?f=Si%7B2%7DCl%7B6%7D%20%28l%29%20%20%2B%20%204H%7B2%7DO%20%28l%29%20%20------%20%3E%202SiO%7B2%7D%20%28s%29%20%20%2B%20%20%20%206HCl%20%28g%29%20%20%20%20%2B%20%20%20%20%20%20H%7B2%7D%20%28g%29)
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Answer:
![K_{p}=4.35\times 10^{-4}](https://tex.z-dn.net/?f=K_%7Bp%7D%3D4.35%5Ctimes%2010%5E%7B-4%7D)
Explanation:
We know, ![K_{p}=K_{c}(RT)^{\Delta n}](https://tex.z-dn.net/?f=K_%7Bp%7D%3DK_%7Bc%7D%28RT%29%5E%7B%5CDelta%20n%7D)
where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and
is difference in sum of stoichiometric coefficient of products and reactants
Here
and T = 311 K
So, ![K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}](https://tex.z-dn.net/?f=K_%7Bp%7D%3D%280.0111%29%5Ctimes%20%5B%280.0821L.atm.mol%5E%7B-1%7D.K%5E%7B-1%7D%29%5Ctimes%20311K%5D%5E%7B-1%7D%3D4.35%5Ctimes%2010%5E%7B-4%7D)
Hence value of equilibrium constant in terms of partial pressure
is ![4.35\times 10^{-4}](https://tex.z-dn.net/?f=4.35%5Ctimes%2010%5E%7B-4%7D)