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Hi, my elder boy ran into an issue with the following.

Hope someone can enlighten him. Many thanks in advance.
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We need to calculate the enthalpy change of a certain reaction, usually acid-base, acid-carbonate, or dissolution reactions. So we take the two substances and mix them together and record the temperature change. Using Q=mcT, we calculate the amount of thermal energy given off or taken in by the reaction, and then using H=Q/n we find the enthalpy change of the reaction. However, the confusion is what value of n (number of moles) to use.

Examples of reactions:

HCl + NaOH -> H20 + NaCl (here n is the mol of any one of the reactants or products, no confusion.)

2HNO3 + Na2CO3 -> H2O + 2NaNO3 (here teacher says take mol of H2O, because it is the one carrying off the thermal energy. Or take it as the mol of Na2CO3, because that is the reactant that is changing, not the acid.)

H2SO4 + KOH -> 2H2O + K2SO4 (here following the same rule as above, taking the number of mol of water, is for some reason wrong.)

Is there a general rule or concept that allows you to figure out exactly which number of mol to use?

Jennifer wrote:Hi, my elder boy ran into an issue with the following.

Hope someone can enlighten him. Many thanks in advance.
******************************************************

We need to calculate the enthalpy change of a certain reaction, usually acid-base, acid-carbonate, or dissolution reactions. So we take the two substances and mix them together and record the temperature change. Using Q=mcT, we calculate the amount of thermal energy given off or taken in by the reaction, and then using H=Q/n we find the enthalpy change of the reaction. However, the confusion is what value of n (number of moles) to use.

Examples of reactions:

HCl + NaOH -> H20 + NaCl (here n is the mol of any one of the reactants or products, no confusion.)

2HNO3 + Na2CO3 -> H2O + 2NaNO3 (here teacher says take mol of H2O, because it is the one carrying off the thermal energy. Or take it as the mol of Na2CO3, because that is the reactant that is changing, not the acid.)

H2SO4 + KOH -> 2H2O + K2SO4 (here following the same rule as above, taking the number of mol of water, is for some reason wrong.)

Is there a general rule or concept that allows you to figure out exactly which number of mol to use?

Hi,

The change in enthalpy equals the amount of heat absorbed or given off by the system during the reaction (at constant pressure).

i.e. H = q (KJ)

As for H = Q/n, it is just more specific in defining the change in enthalpy per mole of which reactant/product reacts/forms.

For example, heat of neutralization/combustion/formation which clearly defined change in enthalpy per mole of "what" reactant/product.

Regards

Oldschool wrote:

Jennifer wrote:the confusion is what value of n (number of moles) to use.

Is there a general rule or concept that allows you to figure out exactly which number of mol to use?

As for H = Q/n, it is just more specific in defining the change in enthalpy per mole of which reactant/product reacts/forms.

For example, heat of neutralization/combustion/formation which clearly defined change in enthalpy per mole of "what" reactant/product.

Thank you for the above.

Is there a general rule or concept that allows you to figure out exactly which number of mol to use?

Jennifer wrote:

Oldschool wrote:

Jennifer wrote:the confusion is what value of n (number of moles) to use.


Is there a general rule or concept that allows you to figure out exactly which number of mol to use?

Which n to use will depend on what is asked.

For example reaction (3), the balanced equation is

H2SO4 + 2KOH = K2SO4 + 2H2O

If Q KJ was given off.

What is the heat evolved per mole of H2SO4 in the reaction?

Then the change in enthalpy per mole of H2SO4 reacts = Q KJ/mol
n=1, because there is 1 mole of H2SO4

However, if the question is changed to "What is the heat of neutralization?"

Then it will be, enthalpy change per mole of
H+ or OH- or H20 = Q/2 KJ/mol
n=2, because there are 2 moles of H+/OH-/H2O

This is because heat of neutralization is defined as the heat change when one mole of H+ from the acid is neutralized by one mole of OH- from an alkali to from one mole of H2O.

Regards

Oldschool wrote:Which n to use will depend on what is asked.

Thank you for the above.

I think he still has some questions. Will post again when he has time.

Hi, for anyone that needs help with lower sec / secondary 3 chemistry, feel free to ask here. Btw, I'm in Sec 3 (O-level).

Can someone help me with this question?
There are 2 unlabelled test-tubes. One contains lime water and the other contains salt solution. Pick one of the following actions to identify the solutions.

1) add sodium hydroxide solution
2) add dilute hydrochloric acid
3) add ammonium carbonate solution
4) add aqueious ammonia

The answer is (3) - can expalin why??
Thanks

When and ammonium compound is added to an aqueous alkali solution (in this case aqueous Ca(OH)2 ), the ammonium salt would react with it to liberate ammonia gas [Ca(OH)2 (aq) + (NH4)2CO3 (aq) -> 2NH3 (g) + 2H2O (l) + CaCO3 (s)] Effervescence of a colourless pungent gas will be observed and a white precipitate is formed. On the other hand, the test tube with the salt solution has a neutral pH, thus is neither an alkali or an acid. Thus, at most a double displacement reaction would occur, but no effervescence would be observed.

Dear hterbin

Thanks for your help.
May I know what is "a double displacement reaction would occur"
Can you explain to me why the other answers are not correct. Thanks a lot.

A displacement reaction can be pictured as the following:

A + BC -> B + AC

It is when an ion displaces another ion of a compound

A double displacement reaction can be pictured as:

AB + CD -> AD + BC

Where the ions of the compounds displaces each other (something like swapping).


In this scenario the answer cannot be sodium hydroxide as the salt solution may contain Grp I or Barium ions. As such rendering it soluble and you can't differentiate.

You also can't use HCl as most chlorides are soluble, which means you can't differentiate the solutions because chances are there would be no precipitate unless it is a Lead or Silver salt solution.

Finally, aqueous ammonia don't react with calcium hydroxide or salt solutions