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    1) why is HI soluble in water?? the compound can only be soluble if the bonds that it will form are stronger than the previous bonds , which are hydrogen bonds. arent the hydrogen bonds stronger than the dipole-dipole forces formed????? so how is it soluble

    2) is hydration energy only for ionic compounds??? or for covalent compounds too??
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    HI separates into H+ and I- ions.

    Look at the definition of hydration enthalpy and you'll see.
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    (Original post by Pigster)
    HI separates into H+ and I- ions.

    Look at the definition of hydration enthalpy and you'll see.
    how does it give ions if its a covalent compound????
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    Exactly the same way as HCl(g) forms ions when it is dissolved.
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    Ionic and covalent compounds are general over-arching names given to chemical compounds; it's more of a continuous scale. So whilst CH4 and HI are both covalent compounds, HI could be said to have more ionic character by virtue of its greater polarity due to greater electronegativity differences between hydrogen and iodine, compared to the difference between hydrogen and carbon.

    You consider solubility, like a lot of things in chemistry, in terms of its Gibbs free energy. This is composed of the enthalpy and entropy term. The entropy is positive as you are going from gas phase HCl are room temperature, to solvated ions, which are more ordered.

    You now consider the enthalpy term; if you were to draw the Hess' cycle and plug in the values, you would see that the process of transforming HI(g) -> H+(aq) + I-(aq) is overall exothermic. The endothermic terms are the bond disassociation enthalpy of H-I, and the ionisation energy of H, whilst the exothermic terms are the first electron affinity of I, as well as the enthalpy of solution of both H+ and I-. The exothermic terms more than compensate for the endothermic terms, resulting in a high solubility. If you were to compare it to say, H-F, which is not very soluble, you would see there are a few things different for its enthalpy terms. The H-F bond is very much stronger than the H-I bond, which is quite weak. Fluorine's first electron affinity is more exothermic than Iodine, and its enthalpy of solution is higher, but still it is less soluble, giving strength to the argument that it is mainly bond enthalpy that decides which is more soluble.
    Considering the overall Gibbs free energy it can be seen that at room temperature, the enthalpic must outweigh the entropic effects.

    Edit: Removed the erroneous statement that the entropy of the reaction is negative
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    (Original post by Razorsharp333)
    Ionic and covalent compounds are general over-arching names given to chemical compounds; it's more of a continuous scale. So whilst CH4 and HI are both covalent compounds, HI could be said to have more ionic character by virtue of its greater polarity due to greater electronegativity differences between hydrogen and iodine, compared to the difference between hydrogen and carbon.

    You consider solubility, like a lot of things in chemistry, in terms of its Gibbs free energy. This is composed of the enthalpy and entropy term. The entropy is of course negative, as there is only one reagent, HI, but two products (H+ and I-).
    You now consider the enthalpy term; if you were to draw the Hess' cycle and plug in the values, you would see that the process of transforming HI(g) -> H+(aq) + I-(aq) is overall exothermic. The endothermic terms are the bond disassociation enthalpy of H-I, and the ionisation energy of H, whilst the exothermic terms are the first electron affinity of I, as well as the enthalpy of solution of both H+ and I-. The exothermic terms more than compensate for the endothermic terms, resulting in a high solubility. If you were to compare it to say, H-F, which is not very soluble, you would see there are a few things different for its enthalpy terms. The H-F bond is very much stronger than the H-I bond, which is quite weak. Fluorine's first electron affinity is more exothermic than Iodine, and its enthalpy of solution is higher, but still it is less soluble, giving strength to the argument that it is mainly bond enthalpy that decides which is more soluble.
    The entropy change is negative as you are going from a gas (high entropy) to solvated ions (relatively low entropy), not becase there are more particles formed. That would suggest a positive entropy change.
 
 
 
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