Course Reader Question

[Alyssa_Hsu_2K]

On pg. 20 of the course reader, there is the following example:
In a constant P calorimeter at 25 Celcius mix 50.0ml each of 1.0 M HCL and 1.0M NaOH. After Mixing, the temperature= 31.9 degrees Celcius.

(H+)+(OH-)--> H2O.

How much heat is released?

In this case, do we assume that all of H and OH- has been converted into water? Is that why we only worry about the specific heat of water?In other words, if we had some H leftover, would we need to take that into account when we do our calculations?

[Lauren Trent 2A]

Since there is an equal amount of a strong acid and a strong base that each have the same molarity, each of the reactants will react entirely as the reaction goes to completion. Therefore all of the H+ and OH- will all convert to water and there will be no excess, which explains why the problem only references the specific heat of water.

[Alyssa_Hsu_2K]

Hi, thank you for your response! However, I'm still a little confused.
On course reader pg. 18, when we have the reaction CH4+2O2--> CO2+2H2O, we take into account the standard enthalpies of both the products and the reactants. However, this particular equation is not at chemical equilibrium, which means CH4 and 2O2 should be completely consumed as well.
So I guess my question is: why do we take reactants into account for CH4+2O2--> CO2+2H2O, but not for HCl+NaOH--> H2O when neither of them is at chemical equilibrium.

Thanks!

[Chem_Mod]

On pg. 20 of the course reader, there is the following example:
In a constant P calorimeter at 25 Celcius mix 50.0ml each of 1.0 M HCL and 1.0M NaOH. After Mixing, the temperature= 31.9 degrees Celcius.

(H+)+(OH-)--> H2O.

How much heat is released?

In this case, do we assume that all of H and OH- has been converted into water? Is that why we only worry about the specific heat of water? In other words, if we had some H leftover, would we need to take that into account when we do our calculations?

We don't need to know if all of H⁺ and ^-OH is converted to water. What we need to know is the temperature change to calculate how much heat is released.

Although, given the large equilibrium constant for this reaction, the equilibrium likes strongly to the right and one could approximate that almost all combine to form water molecules.

In our thermodynamic problems we will calculate changes as per the balanced equation or for a specific given amount of reactant, and assume all reactant goes to product (unless there is a limiting reactant). See HW problems.

[Alyssa_Hsu_2K]

Hello,
I was wondering why we don't need to take into account the temperature change that may arise when reactants break their bonds.
Thanks!