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### Topic: Quick Enthalpy Change Question  (Read 6295 times)

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#### idiotatchemistry

• Guest
##### Quick Enthalpy Change Question
« on: February 22, 2005, 12:25:07 PM »
I'm calculating the enthalpy change of a reaction in solution.

KHCO3 (s) + HCl(aq) -> KCL (aq) + H2O (l) + CO2 (g)

Basically I used 30cm^3 of 2Mol dm^-3 HCl acid and put in 3.9g of potassium hydrogencarbonate.
I recorded the temp change, it decreased at the end of 7minutes by 7C, this would mean it is an endothermic reaction and so whatever the enthalpy change is, will be a postive number, correct?

Now that I want to work out the Heat given out by the solution, for mass do I use 30g (assuming 30cm^3 of the HCl acid is 30g) + 3.9g?
So should the mass I use to calculate the heat given out by 33.9g, or should it just be thirty?

Assuming I should use 33.9g, and knowing the spec. heat capacity is 4.2 and the temp change is -7...the heat given out should be 997J, correct?

The next bit confuses me. I am supposed to use whatever value I got for Heat given out and divide that by a concentration, so it should be
997J/2 (conc of the HCl acid) to get the enthalpy change of this reaction, right? Or have I got it wrong anywhere above?

Thx

#### Demotivator

• Guest
##### Re:Quick Enthalpy Change Question
« Reply #1 on: February 22, 2005, 06:51:23 PM »
1. correct, endothermic
2. just 30g  for the  mass because the medium that carries the heat is the solvent  for which the specific heat is given.
The calculation is otherwise correct.

3. First the limiting reagent needs to be determined since the quantities of reactants are not necessarily stoichiometric. I suspect its the bicarbonate. Then determine moles of sodium bicarbonate if limiting reagent and divide into heat to get  Joules/mole sod bicarb.  It is not conventional to divide by concentration, ie moles/L unless you're sure that is what they want.