# Enthalpy Change Help

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#1
-I've used experimental data to calculate the enthalpy change for the reaction of potassium carbonate and potassium hydrogen carbonate with hydrochloric acid. These were the equations I was given:

When working out the enthalpy change for this equation:

Do I times the potassium hydrogen carbonate value by 2 and keep this value or divide it by 2 once I've calculated the final enthalpy change. Or do I ignore the multiplying. This is something I'm just conceptually confused about as I'm using my own experimental data.
Last edited by Obolinda; 8 months ago
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8 months ago
#2
which reaction do you have data for
1
#3
(Original post by mechengineer8)
which reaction do you have data for
reaction 1 and 2 to find the enthalpy change for reaction 3
I know how to do the hess diagram, its just because reaction 3 has coefficient of 2 on the reactant so I'm wondering if I still need to multiply the enthalpy change of reaction 2 by 2
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8 months ago
#4
Use Hess Law

Times what you got for KHCO3 by 2, yes.
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#5
(Original post by CaptainDuckie)
Use Hess Law

Times what you got for KHCO3 by 2, yes.
oh ok thanks
i was just worried as the enthalpy changes I've got was something I calculated through a calorimetry experiment
1
8 months ago
#6
Enthalpy is a state function.

This means if you have an equation like this:

2KHCO3 --> K2CO3 + CO2 + H2O and ΔH = +8.5 kJmol-1

Then if you double the equation then ΔH also doubles. Likewise, if you reverse reactants with products, ΔH is the same but negative.

4KHCO3 --> 2K2CO3 + 2CO2 + 2H2O ΔH = +17 kJmol-1

2KHCO3 --> K2CO3 + CO2 + H2O --> 2KHCO3 ΔH = -8.5 kJmol-1

If you multiply reaction 2 by 2 and then add the reverse of reaction 1 in your question (bearing in mind the effect on ΔH shown above) you can calculate ΔH. (Cancel out same things on left and right side of equation when you add the equations together).

Hope that helps.
1
#7
(Original post by MoleChem)
Enthalpy is a state function.

This means if you have an equation like this:

2KHCO3 --> K2CO3 + CO2 + H2O and ΔH = +8.5 kJmol-1

Then if you double the equation then ΔH also doubles. Likewise, if you reverse reactants with products, ΔH is the same but negative.

4KHCO3 --> 2K2CO3 + 2CO2 + 2H2O ΔH = +17 kJmol-1

2KHCO3 --> K2CO3 + CO2 + H2O --> 2KHCO3 ΔH = -8.5 kJmol-1

If you multiply reaction 2 by 2 and then add the reverse of reaction 1 in your question (bearing in mind the effect on ΔH shown above) you can calculate ΔH. (Cancel out same things on left and right side of equation when you add the equations together).

Hope that helps.
thanks !
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