Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: iyc on November 10, 2005, 05:25:37 AM
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In determination of enthaply change of hydration of MgSO4:
MgSO4(s) + 7H2O -> MgSO4.7H2O(s)
It is impossible to measure the enthalpy change for this reaction. But, I don't know why?
Is it due to the activation energy of the reaction too high that cannot be overcome under standard conndition?
Besides, when M g of anhydrous magnesium sulphate(VI) were add to V cm3 of water in a beaker and all magnesium sulphate is dissolved, there was a maximum rise in temperature of the solution by ToC. When calculating the enthalpy evolved by using E=mc (deltaT), what should I substitute for mass? Is mass equal to the sum of mass of anhydrous magnesium sulphate(VI) and mass of water added?
Thank you.
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In determination of enthaply change of hydration of MgSO4:
MgSO4(s) + 7H2O -> MgSO4.7H2O(s)
It is impossible to measure the enthalpy change for this reaction. But, I don't know why?
Is it due to the activation energy of the reaction too high that cannot be overcome under standard conndition?
Besides, when M g of anhydrous magnesium sulphate(VI) were add to V cm3 of water in a beaker and all magnesium sulphate is dissolved, there was a maximum rise in temperature of the solution by ToC. When calculating the enthalpy evolved by using E=mc (deltaT), what should I substitute for mass? Is mass equal to the sum of mass of anhydrous magnesium sulphate(VI) and mass of water added?
Thank you.
you probably want to work with that of water
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I would assume that you have to use Hess' Law to determine the enthalpy of hydration. If you added MgSO4 to water I think you would just be measuring the enthalpy for:
MgSO4 ------> Mg2+ + SO42-
Were you given any equations with corresponding enthalpy changes?
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In determination of enthaply change of hydration of MgSO4:
MgSO4(s) + 7H2O -> MgSO4.7H2O(s)
It is impossible to measure the enthalpy change for this reaction directly. But, I don't know why?
Is it due to the activation energy of the reaction too high that cannot be overcome under standard conndition?
In fact, I would like to ask this part.
I think the activation energy of the reaction is too high that cannot be overcome under standard condition. That means, the the reactants are kinetically stable.Is it right if I say so?
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Well how would you measure it directly?
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Can I measure enthalpy change of this reaction MgSO4(s) + 7H2O -> MgSO4.7H2O(s) directly?
If not, is it due to the high activation energy that cannot be overcome spontaneously under standard condition?
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MgSO4(s) + 7H2O -> MgSO4.7H2O(s)
I don't think you can measure this directly. Just the practicality of it I think. If you just took MgSO4 and put it in excess water in a calorimeter you would just be measuring the dissociation of MgSO4 I think.
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Thank you, I think I can solve the problem.