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Topic: Hess Law: Molar Enthalpy Change for Decomposition  (Read 17492 times)

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Offline Tomtom

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Hess Law: Molar Enthalpy Change for Decomposition
« on: April 11, 2007, 12:40:33 PM »
1. The problem statement, all variables and given/known data
This is for a chemistry report. This is the task:
"Determine the molar enthalpy change for the decomposition of sodium hydrogen carbonate into sodium carbonate, CO(2) and water.
2NaHCO3 -> Na2CO3 + CO2 + H2O
This enthalpy change cannot be measured directly".

Instead, we reacted NaHCO3 and Na(2)CO(3) in separate containers with HCL (in excess (100 mL), about 1M).

We measured the temperature changes, and the masses of each solid/liquid.

For the NaHCO3 reaction:
Mass(NaHCO3): 0.001021 kg ± 0.000001 kg
Mass(HCl): 0.100 ± 0.001 kg
Change in T = 1 K

For the Na2CO3 reaction:
Mass(Na2CO3): 0.000905 kg ± 0.000001 kg
Mass(HCl): 0.100 ± 0.001 kg
Change in T = -0.3 K

2. Relevant equations
Nr 1: 2NaHCO3 -> Na2CO3 + CO(2) + H2O
Nr 2: NaHCO3 + HCl -> NaCL + CO2 + H2O
Nr 3: Na2CO3 + HCl -> 2NaCL + CO2 + H2O

Note: By multiplying equation 2 by two, reversing equation 3, and then adding them together, we get equation 1. This has to be done to the

3. The attempt at a solution
My biggest problem is figuring out how to find the Molar Delta H in kJ/mol. By doing the following, I can calculate the approximate Delta H (only in kJ) for reaction 1:

Following the equation of Heat = Mass * Specific Heat (c) * Delta T, we get:

(Note, here I add the two masses of HCl and the solids, and then use the specific heat of water. I understand this holds quite a large uncertainty. If anybody has a better idea, please say so. I am also really unsure on what I do after calculating the heat, ie. when I invert the sign of the heat form + to - and vica versa. Is this correct?)

For NaHCO3:
Mass = 0.101021 kg
Specific Heat (water) 4186 J/(kg * K)
Delta T = 1 K

Heat = 422.87 J. As this is an exothermic reaction, it becomes -422.87 J

For Na2CO3:
Mass = 0.100905 kg
Delta T = -0.3 K

Heat = -126.71 J. As this is an endothermic reaction, it becomes +126.71 J.

Then, by doing the same to these numbers as is mentioned in Part 2 (the reactions), I end up with (-422.87 * 2) + (126.71*-1) = -970.71 J = -0.971 kJ!

But this however, is just the enthalpy change for the reaction. Not the molar enthalpy change...
Any comments on this would be absolutely wonderful, and greatly appreciated!

allanf

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Re: Hess Law: Molar Enthalpy Change for Decomposition
« Reply #1 on: April 11, 2007, 07:53:06 PM »
(Note, here I add the two masses of HCl and the solids, and then use the specific heat of water. I understand this holds quite a large uncertainty. If anybody has a better idea, please say so. I am also really unsure on what I do after calculating the heat, ie. when I invert the sign of the heat form + to - and vica versa. Is this correct?)

It sounds like you are multiplying the mass of reagents (HCl and the carbonates) and the heat capacity (specific heat) of water together. (it should be product of the mass of water by the heat capacity of water)

You should convert the masses of the solids, and the amount of HCl used in each case, to moles. Then you can determine the limiting reagent.  The molar enthalpy of reaction for each decomposition reaction is then just the total heat measured divided by the moles of limiting reagent.

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