lhh2003
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For 13)b)ii), how are we meant to know that the reaction is reversible and that we should think about the reaction in terms of Le Chatelier's principle ?

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The enthalpy change of the reaction of dissolving CaCl2 is Exothermic.

Surely it can be argued a higher temperature would increase solubility as H2O molecules have more KE so collide with molecules ions more to hydrate them more, giving a more exothermic enthalpy change of solution, therefore, increased solubility ? Where am I going wrong ?

Thanks.
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charco
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(Original post by lhh2003)
For 13)b)ii), how are we meant to know that the reaction is reversible and that we should think about the reaction in terms of Le Chatelier's principle ?

This is what the question is :

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Views: 35
Size:  111.0 KB

And here is the mark scheme :


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The enthalpy change of the reaction of dissolving CaCl2 is Exothermic.

Surely it can be argued a higher temperature would increase solubility as H2O molecules have more KE so collide with molecules ions more to hydrate them more, giving a more exothermic enthalpy change of solution, therefore, increased solubility ? Where am I going wrong ?

Thanks.
You are confusing kinetics with thermodynamics.

Yes, increased temperature will increase the RATE of dissolution as the particles are moving faster.

BUT, the saturation point is when no more solute dissolves and effectively you are in equilibrium between:
solute(s) <==> solute (aq)

Now you have to consider the thermodynamics of the effect of increased energy on the equilibrium (actually there are kinetics involved, but lets not get too technical)

According to Le Chatelier's principle, a system at equilibrium will respond to any change in conditions in such a way as to "cancel out" the change. If you increase the temperature the system moves to the endothermic direction, absorbing heat.

The endothermic direction in this case reduces the solubility.
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lhh2003
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(Original post by charco)
You are confusing kinetics with thermodynamics.

Yes, increased temperature will increase the RATE of dissolution as the particles are moving faster.

BUT, the saturation point is when no more solute dissolves and effectively you are in equilibrium between:
solute(s) <==> solute (aq)

Now you have to consider the thermodynamics of the effect of increased energy on the equilibrium (actually there are kinetics involved, but lets not get too technical)

According to Le Chatelier's principle, a system at equilibrium will respond to any change in conditions in such a way as to "cancel out" the change. If you increase the temperature the system moves to the endothermic direction, absorbing heat.

The endothermic direction in this case reduces the solubility.
Ahh ok, I see what you are saying.

Just another point, doesn't the definition of enthalpy change refer to the energy change per ONE MOLE of solute that DISSOLVES ; not the energy change of ATTEMPTING TO dissolve one mole of solute and see the energy change ?
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charco
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(Original post by lhh2003)
Ahh ok, I see what you are saying.

Just another point, doesn't the definition of enthalpy change refer to the energy change per ONE MOLE of solute that DISSOLVES ; not the energy change of ATTEMPTING TO dissolve one mole of solute and see the energy change ?
The actual definition of enthalpy of solution is quite tricky.

It is the energy change when 1 mol of a solute dissolves in an "infinite" volume of solvent.

Clearly, this is not to be taken literally. It just means that enough volume of solvent is used so that if any greater amount of solvent were to be used there would be no further enthalpy change.
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