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There are two separate experiments done in this problem. First experiment

106.6 gram of water at 23.28 C and 8.56 gram of sodium sulfate places into a foam cup (a calorimeter) the temperature to 24.97C. The product are sodium sulfate decahydrate.

Second experiment

105.4 g of water at 24.4C and 19.3 grams of sodium sulfate decahydrate. The temp this time rises lowers to 17.34.(the question didn't state what the product is of this reaction, but I guess it's Na2SO3•20H2O)

Note the specific heat of both solutions is that of water-4.18. And that heat of reaction is -heat of calorimeter

The question is what the molar enthalpy of hydration of sodium sulfate is. also one more questionhow to determine the theoretical value?(after yomen answered my question I realized it is basically a reaction from sodium sulfate 10-hydrate to 20 hydrate as product, if you were to combine the two reaction as one(Hess Law)but the thing is i can't find the heat of formation of any hydrate compounds!?)

Now the question refers to the first experiment, so I multiply the the total mass of water of and sodium sulfate by the specifc heat of water and then multiply the product by the change in temperature and I got 813.513J

Because this is the enthalpy for 0.060263 moles of sodium sulfate so Divide 813.513 by 0.060263 and I got 13499.37J.

Before starting solving for the value of molar enthalpy I knew I am probably gonna get a wrong answer because:

  1. By doing this way, I render the second experiment completely useless.

  2. The final answer I knew I was gonna get is way too big. enter image description here

(Chances are I may read the question wrong...).

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  • $\begingroup$ What you did for the first experiment is quite accurate. But I don't agree that for the second experiment you will have Na2SO3•20H2O. $\endgroup$ – Yomen Atassi Jan 3 '15 at 14:21
  • $\begingroup$ What will it come out it be? $\endgroup$ – most venerable sir Jan 3 '15 at 15:04
  • $\begingroup$ @yomen, please help me correct my understanding. $\endgroup$ – most venerable sir Jan 3 '15 at 15:07
  • $\begingroup$ Honestly, I don't know. I'm searching... $\endgroup$ – Yomen Atassi Jan 3 '15 at 19:42
  • $\begingroup$ It shouldn't be too hard though, I can give you a photo of the original problem in case I copy it wrong. $\endgroup$ – most venerable sir Jan 3 '15 at 19:43
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Enthalpy of hydration is the energy change for converting 1 mol of an anhydrous substance to 1 mol of the hydrated substance. In order to find this number, it is necessary to first calculate the enthalpy of dissolution for each substance separately, and then find the difference between the two. The enthalpy of dissolution is the energy change of dissolving 1 mol of a substance in water. It is calculated using temperature changes in the water, heat capacity of the substance, and the weight of the mixture.

-So, what you have calculated from the first experiment is the heat of dissolution of the anhydrous sodium sulfate.

-You have to repeat the calculation for the second experiment to determine the heat of dissolution of hydrated sodium sulfate.

  • The enthalpy of hydration is the difference between the two heats of dissolution. It's always negative.
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  • $\begingroup$ Thank you. This is of great help! May I know where you found the info? $\endgroup$ – most venerable sir Jan 3 '15 at 21:38
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    $\begingroup$ adamcap.com/schoolwork/… $\endgroup$ – Yomen Atassi Jan 3 '15 at 21:41
  • $\begingroup$ so I guess it is basically from 10 hydrate sodium sulfate to 20 hydrate sodium sulfate. $\endgroup$ – most venerable sir Jan 4 '15 at 0:33
  • $\begingroup$ By Hess law I guess $\endgroup$ – most venerable sir Jan 4 '15 at 0:33
  • $\begingroup$ How to find the theoretical value? How to find the heat of formation of something with hydrate ? $\endgroup$ – most venerable sir Jan 4 '15 at 0:50

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