Practice Problem: Hess's Law
Summary
TLDRThis educational video script explains Hess's Law, a fundamental concept in thermodynamics, using a step-by-step approach to solve for the enthalpy change of a reaction. The script guides viewers through manipulating given thermochemical equations by adjusting stoichiometric coefficients and enthalpy values, ensuring that unwanted substances cancel out. The process involves halving coefficients and enthalpy changes, flipping reactions, and adding the modified enthalpy values to find the desired ΔH for the reaction ClF + F2 → ClF3. The final result is a clear demonstration of how to apply Hess's Law to determine the enthalpy change for a chemical reaction.
Takeaways
- 🔍 Hess's Law is used to calculate the change in enthalpy for a reaction when direct measurement is not possible.
- 📐 The given thermochemical data includes three reactions with known enthalpy changes.
- 🔄 To use Hess's Law, manipulate the given equations by reversing or adjusting coefficients to match the desired reaction.
- ✂️ Halving the coefficients in an equation also halves the corresponding enthalpy change (ΔH).
- 🔀 Reversing a reaction changes the sign of ΔH and adjusts the stoichiometric coefficients to match the desired direction.
- 🧩 Combine the manipulated equations to achieve the target reaction, ensuring that unwanted substances cancel out.
- 📉 When combining ΔH values, add them together based on their manipulated signs and magnitudes.
- 🌡️ The final ΔH for the target reaction is calculated by summing the adjusted ΔH values from the manipulated equations.
- ⚖️ Ensure that the reactants and products balance out correctly after combining the equations.
- 📝 The process involves careful manipulation of equations and ΔH values to isolate the desired reaction.
Q & A
What is Hess's Law and how is it used in the context of the given problem?
-Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps the reaction is carried out in. In the context of the problem, it is used to calculate the enthalpy change for a reaction where the direct enthalpy change is unknown by combining known enthalpy changes from other related reactions.
Why is it necessary to manipulate the stoichiometric coefficients and the ΔH values when applying Hess's Law?
-Manipulating the stoichiometric coefficients and ΔH values is necessary to align the reactants and products of the given reactions with the desired target reaction. This allows for the correct combination of the reactions to achieve the desired outcome, ensuring that unwanted substances cancel out and the target reaction is isolated.
How does the process of 'flipping' a reaction contribute to solving the problem?
-Flipping a reaction means reversing the direction of the reaction, which changes the products to reactants and vice versa. This is crucial when the desired species are present in the wrong state (i.e., as a reactant when needed as a product, or vice versa). By flipping, the reactions can be adjusted to match the required form for the target reaction.
What is the significance of cutting the stoichiometric coefficients and ΔH values in half in the given example?
-Cutting the stoichiometric coefficients and ΔH values in half is done to adjust the amount of a substance in a reaction to match the target reaction's requirement. In the example, ClF is needed in a 1:1 ratio, but the given reaction has it in a 2:1 ratio, so halving the reaction ensures the correct stoichiometry for the calculation.
Why is it important to ensure that unwanted substances cancel out when using Hess's Law?
-Ensuring that unwanted substances cancel out is important because it allows for the isolation of the desired reaction's enthalpy change. If substances do not cancel out, they would incorrectly contribute to the overall enthalpy change of the reaction, leading to an incorrect result.
How does the process of adding up the ΔH values from the manipulated reactions yield the enthalpy change for the target reaction?
-By correctly manipulating and aligning the given reactions to match the target reaction, the ΔH values from each step can be added together. Since enthalpy changes are state functions, the sum of the enthalpy changes from the individual steps will equal the enthalpy change for the overall reaction.
What is the final ΔH value calculated for the reaction ClF + F2 → ClF3 in the given example?
-The final ΔH value calculated for the reaction ClF + F2 → ClF3 is -139.2 kilojoules.
Why is it necessary to flip the sign of the ΔH value when flipping a reaction?
-Flipping the sign of the ΔH value is necessary because the enthalpy change is a measure of the energy absorbed or released by the system. When a reaction is reversed, the energy change associated with it is also reversed, thus the sign of ΔH must be flipped to reflect this change.
Can Hess's Law be used to determine the enthalpy change for any chemical reaction?
-Hess's Law can be used to determine the enthalpy change for any chemical reaction where the individual steps and their enthalpy changes are known. It is particularly useful when direct measurement of the enthalpy change for a reaction is not feasible.
What is the role of the tutorial mentioned in the script in understanding Hess's Law?
-The tutorial mentioned in the script serves as a supplementary resource for those unfamiliar with Hess's Law. It provides a detailed explanation of the concepts and techniques needed to solve problems like the one presented in the script.
Outlines
🔍 Hess's Law Application
This paragraph introduces the application of Hess's Law to determine the enthalpy change for a chemical reaction. The video script discusses how to use given thermochemical data for three related reactions to find the enthalpy change for the reaction ClF + F2 → ClF3. The narrator guides viewers on how to manipulate the equations by halving certain reactants and products, flipping reactions, and adjusting the enthalpy values accordingly. The process involves identifying the required substances in the given equations, modifying the stoichiometric coefficients, and ensuring that unwanted substances cancel out in the final equation.
🧮 Calculation of Enthalpy Change
The second paragraph continues the explanation of using Hess's Law by detailing the calculation of the enthalpy change for the desired reaction. The narrator ensures that all reactants and products are correctly aligned and that they cancel out appropriately in the final equation. The focus is on ensuring that the final equation matches the target reaction. The enthalpy values from the manipulated equations are then added together to find the overall enthalpy change for the reaction ClF + F2 → ClF3, which is calculated to be -139.2 kilojoules.
Mindmap
Keywords
💡Hess's Law
💡Thermochemical Data
💡Enthalpy Change (ΔH)
💡Stoichiometric Coefficients
💡Reactants and Products
💡Flipping Reactions
💡Cutting in Half
💡Canceling Out
💡Kilojoules (kJ)
💡Chemical Equations
Highlights
Introduction to Hess's Law and its application in thermochemical calculations.
Given thermochemical data for three reactions to find the enthalpy change for ClF + F2 → ClF3.
Explanation of how to use Hess's Law to solve for unknown enthalpy changes.
Identification of the need for ClF in the second given equation.
Adjustment of the second equation by halving the stoichiometric coefficients and enthalpy change.
Requirement for F2 as a reactant and its presence in the first equation as a product.
Flipping and halving the first equation to align reactants and products correctly.
Adjustment of the enthalpy change when flipping the equation sign and halving the value.
Identification of ClF3 in the third equation as a reactant.
Flipping the third equation to have ClF3 as a product.
Ensuring that unwanted substances cancel out in the combined reactions.
Cancellation of O2, Cl2O, and OF2 across the reactions to isolate the desired reaction.
Final assembly of the desired reaction ClF + F2 → ClF3 using the manipulated equations.
Summation of the enthalpy changes from the manipulated equations to find ΔH for the desired reaction.
Result of the calculation: ΔH = -139.2 kJ for the reaction ClF + F2 → ClF3.
Emphasis on the importance of correctly manipulating and aligning the given data with the desired reaction.
Transcripts
okay let's try a question about Hess's law. so given the thermochemical data
below those are the three reactions at the bottom we want to find the change in
enthalpy for the following reaction so we have ClF + F2 yields ClF3
and that Delta H value is unknown but we have three somewhat related equations at
the bottom with known thermo chemical data see how we have those Delta H's
there so we can use a technique called Hess's law to solve this problem and if
you're unfamiliar go ahead and check out my tutorial on Hess's law that will
explain everything you need to know to solve this problem and when you're ready give it a try.
ok so up top we have what we need to figure out and then on the
bottom we have all of our given data that's the stuff that we can use that's
what we can work with and manipulate in order to get the Delta H that we want. so
where are we going to start? let's start out by looking at what we need in our
given equation and then finding where those substances are in the given
equations so we know we need ClF we need ClF we need to find where that is in
these equations down here well the second equation has 2ClF so there's
that whole equation but it has ClF in it that's the only place we have it so
we know we need to use this equation the problem is we have to ClF we only
want one ClF so what should we do well what we can do is we can cut it in half
we're gonna cut it in half meaning we can cut the stoichiometric coefficients
in half and we will therefore also cut the Delta H value in half so we've done
this here 2ClF becomes ClF, O2 becomes 1/2 O2, Cl2O becomes
one half Cl2O, and OF2 becomes 1/2 OF2. now we mustn't forget to cut that
Delta H in half as well because we're cutting all the stoichiometric
coefficients in half that means that Delta H will be half as much as well so
two O five point six becomes one oh two point eight. now what's next we need F2
we need F2 as a reactant so where's F two well over here we have 2F2 and we
have it as a product and that's the only place F2 is so we know we need to use
this equation so what are we going to do we need F2 as a reactant but here we
have 2F2 as a product so what we need to do is we need to take this first
reaction we have to flip it so that the products become reactants and reactants
become the products and we also have to cut it in half so we could do two
manipulations here so let's do that products become the reactants. so O two
as a product becomes one half O2 as a reactant and then 2F2 as a product
becomes F2 as a reactant and then 2OF2 as a reactant becomes one OF2 as
a product so we have flipped it and we have cut it in half. and then don't
forget we have to also manipulate the Delta H value in the corresponding way
so when we flip it whenever we make the products the reactants in the reactants
to the products all we have to do is flip the sign so we took our negative
forty nine point four and we made it positive forty nine point four but we
also cut everything in half so from positive forty nine point four we went
down to positive 24 point seven so that is the Delta H value for this reaction
and lastly we need ClF3 and we need it as a product so where can
we find it here well this third reaction has ClF3 but it has it as a reactant so
what can we do all we have to do is flip the third reaction so let's do that now
so one-half Cl2O as a product becomes a reactant
three-halves OF2 becomes a reactant and then ClF3 which was reactant is
now a product and O2 is also now a product and once
again we can't forget we have to take our Delta H value it was positive to
sixty six point seven because we flipped it it is now negative to sixty six point
seven kilojoules so we've got everything we need so let's just check and make
sure what cancels out let's make sure everything that we don't want in there
is going to cancel out so what do we have well on the Left we
have a half O2 in the first reaction we have a half O2 in the second reaction
those are both reactants and then we have O2 as a product and the third
reaction so half and half makes one and then we have one on the right so O2's
all cancel out. then we have one-half Cl2O
as a product in the first reaction and we have one-half Cl2O
as a reactant in the third reaction so those will also cancel out nicely then
on the right we have one half OF2 in the first reaction we have one OF2 in
the second reaction and then in the third reaction we have three halves OF2
as a reactant so three halves OF2 total as product and three halfs
as reactants and so those will also cancel nicely and what are we left with
we are left with just what we wanted we've got ClF we've got F2 and we've
got ClF3 so that is what we wanted we do end up with the reaction as listed at
the top so all we have to do is now add up the corresponding enthalpy data and
so for the first one we've got one or two point eight for the second one we've
got 24 point seven and for the third one we've got negative two sixty six point
seven so we put those in the calculator add them up and we end up with negative
one hundred thirty nine point two kilojoules that will be the Delta H for the reaction in question.
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