# Enthelpy of Reaction (Hess's Law)

• mrxtothaz
In summary, the provided lab results show three reactions: Reaction 1, Reaction 2, and Reaction 3. The enthalpies for each reaction are -5.80 kJ, -5.06 kJ, and -16.7 kJ, respectively. The question asks to add Reaction 1 and Reaction 2 using Hess's Law and compare the sum with Reaction 3. The attempt at a solution shows that the equations for Reactions 1 and 2 are added to yield the same equation as Reaction 3. However, the enthalpies are significantly different. The only difference between the two equations is the presence of a hydroxide ion on both sides of the equation in the addition of Reactions 1
mrxtothaz

## Homework Statement

I have been provided with the results of a lab:

Reaction 1:
NaOH(s) --> Na+(aq) + OH-(aq)
Reaction 2:
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) --> H2O(l) + Na+(aq) + Cl-(aq)
OH-(aq) + H+(aq) --> H2O(l)
Reaction 3:
NaOH(s) + H+(aq) + Cl-(aq) --> H2O(l) + Na+(aq) + Cl-(aq)
NaOH(s) + H+(aq) --> H2O(l) + Na+(aq)

The enthalpies for each reaction:
∆H°r: -5.80 kJ (Reaction 1)
∆H°r: -5.06 kJ (Reaction 2)
∆H°r: -16.7 kJ (Reaction 3)

I have been asked to add the equations of 1 and 2 (Hess's Law), and to compare the sum with equation 3.

Hess's Law.

## The Attempt at a Solution

NaOH(s) --> Na+(aq) + OH-(aq) ∆H°r: -5.80 kJ (Reaction 1)
OH-(aq) + H+(aq) --> H2O(l) ∆H°r: -5.06 kJ (Reaction 2)

NaOH(s) + H+(aq) + OH-(aq) --> Na+(aq) + H2O(l) + OH-(aq) ∆H°r: -10.9 kJ (Reaction 1+2 intermediate)

NaOH(s) + H+(aq) --> Na+(aq) + H2O(l) ∆H°r: -10.9 kJ (Reaction 1+2)

NaOH(s) + H+(aq) --> Na+(aq) + H2O(l) ∆H°r: -16.7 kJ (Reaction 3)

So, as can be seen, the equations for reactions 1&2 are added and are identical to that of reaction 3. However, the enthalpies of reaction are way different. I do not know how I can account for this steep a difference, as all the lab results were provided with the question. The only difference is that, in the addition of reactions 1&2, there was a hydroxide ion on both sides of the equation that was canceled out (since I was asked for the net ionic equations for each reaction). Could this account for the difference, or does it have no role in the enthalpy of reaction, given that it is a spectator ion?

If I did nothing wrong, I suspect I am supposed to identify this discrepancy between reaction 1+2 with reaction 3.

I'm looking into possible sources of error; the only thing I can think of is that the experimenters assumed the specific heat capacity of each solution to be same as that of water? As with density.

Could this account for the discrepancy?

mrxtothaz said:

## Homework Statement

I have been provided with the results of a lab:

Reaction 1:
NaOH(s) --> Na+(aq) + OH-(aq)
Reaction 2:
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) --> H2O(l) + Na+(aq) + Cl-(aq)
OH-(aq) + H+(aq) --> H2O(l)
Reaction 3:
NaOH(s) + H+(aq) + Cl-(aq) --> H2O(l) + Na+(aq) + Cl-(aq)
NaOH(s) + H+(aq) --> H2O(l) + Na+(aq)

The enthalpies for each reaction:
∆H°r: -5.80 kJ (Reaction 1)
∆H°r: -5.06 kJ (Reaction 2)
∆H°r: -16.7 kJ (Reaction 3)

I have been asked to add the equations of 1 and 2 (Hess's Law), and to compare the sum with equation 3.

Hess's Law.

## The Attempt at a Solution

NaOH(s) --> Na+(aq) + OH-(aq) ∆H°r: -5.80 kJ (Reaction 1)
OH-(aq) + H+(aq) --> H2O(l) ∆H°r: -5.06 kJ (Reaction 2)

NaOH(s) + H+(aq) + OH-(aq) --> Na+(aq) + H2O(l) + OH-(aq) ∆H°r: -10.9 kJ (Reaction 1+2 intermediate)

NaOH(s) + H+(aq) --> Na+(aq) + H2O(l) ∆H°r: -10.9 kJ (Reaction 1+2)

NaOH(s) + H+(aq) --> Na+(aq) + H2O(l) ∆H°r: -16.7 kJ (Reaction 3)

So, as can be seen, the equations for reactions 1&2 are added and are identical to that of reaction 3. However, the enthalpies of reaction are way different. I do not know how I can account for this steep a difference, as all the lab results were provided with the question. The only difference is that, in the addition of reactions 1&2, there was a hydroxide ion on both sides of the equation that was canceled out (since I was asked for the net ionic equations for each reaction). Could this account for the difference, or does it have no role in the enthalpy of reaction, given that it is a spectator ion?

If I did nothing wrong, I suspect I am supposed to identify this discrepancy between reaction 1+2 with reaction 3.

How did you get H+aq? From HAaq?

chemisttree said:
How did you get H+aq? From HAaq?

Not sure what you mean by HAaq, but the hydrogen ions arise from the dissolution of hydrochloric acid.

Then shouldn't there be an enthalpy of HClaq -----> H+aq + Cl-aq?

I'm sure a solution can be found by those means, but the three reactions should be enough for this kind of exercise.

The first two equations combined yield the third equation, so there is no need for additional equations. The only issue is that the enthalpies are quite a bit off.

Do you believe that this difference can be accounted for by the fact that density & heat capacity of water was assumed for each solution? Or is there an error in any of my work?

Note that no experimental error can be attributed to my doing, since the data itself was provided with the question.

## What is Hess's Law?

Hess's Law states that the enthalpy change of a chemical reaction is the same regardless of the route taken from reactants to products. This means that the enthalpy change is a state function and only depends on the initial and final state of the reaction, not the pathway in between.

## What is the enthalpy of reaction?

The enthalpy of reaction, also known as the heat of reaction, is the amount of heat released or absorbed during a chemical reaction. It is typically measured in kilojoules per mole (kJ/mol) and is a way to quantify the energy changes that occur during a chemical reaction.

## How is the enthalpy of reaction calculated?

The enthalpy of reaction can be calculated using Hess's Law and a series of known enthalpy values for different reactions. The enthalpy of reaction is equal to the sum of the enthalpy of formation of the products minus the sum of the enthalpy of formation of the reactants.

## Why is Hess's Law important in chemistry?

Hess's Law is important because it allows us to calculate the enthalpy of a reaction even if we cannot directly measure it. It also helps us understand the relationship between different reactions and how they contribute to the overall energy changes in a system.

## What are some real-life applications of Hess's Law?

Hess's Law has many practical applications, such as in industries that involve energy production or storage. It is also used in environmental studies to understand the energy changes in natural processes, such as combustion reactions in the atmosphere. Additionally, Hess's Law is used in the development of new materials and fuels by predicting their energy changes during reactions.

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