HSC Chemistry: Measure the enthalpy of Neutralisation| Module 6
Summary
TLDRThis script discusses the principles of calorimetry, focusing on measuring the enthalpy change of reactions. It explains the setup of a typical calorimeter experiment, introduces the concept of specific heat capacity, and demonstrates how to calculate enthalpy change using the formula q = mcΔT. The video also covers molar enthalpy of neutralization, highlighting the standard value of -57 kJ/mol for strong acid-strong base reactions and noting that weaker acids or bases result in less energy release. Practical aspects, such as choosing the right base for neutralizing a sulfuric acid spill, are also addressed, emphasizing safety and the importance of minimizing heat loss in experiments.
Takeaways
- 🔍 Calorimetry is the process of determining the enthalpy change of a reaction by measuring the heat absorbed or released.
- 🧪 A calorimeter is an instrument used to measure the enthalpy change in a system, often involving a setup with an insulated cup and a thermometer.
- 🌡️ The temperature change during a reaction is a key measurement, which can be used to calculate the enthalpy change when combined with the mass of the substance and its specific heat capacity.
- ⚖️ The specific heat capacity (c) is the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius or Kelvin, typically 4.18 J/g·K for water.
- 🔄 The enthalpy change (ΔH) is calculated using the formula ΔH = -q/n, where q is the heat exchanged and n is the number of moles of product formed.
- 💧 Molar enthalpy of neutralization is the heat change when one mole of water is formed in a neutralization reaction, typically around -57 kJ/mol for strong acid-strong base reactions.
- 🌡️ Lab conditions for measuring molar enthalpy of neutralization are standardized at 25 degrees Celsius and 100 kilopascals to ensure consistency.
- 🧩 In a neutralization reaction, all reactions go to completion regardless of the strength of the acid or base, but the enthalpy change can vary with the presence of weak acids or bases.
- 🚨 In chemical spills, using a weak base like sodium carbonate can be safer than a strong base due to less heat release and its ability to absorb the acid.
- 📝 The importance of significant figures in recording measurements and calculations cannot be overstated, as incorrect rounding can lead to significant errors in results.
- 🔍 The main source of error in calorimetry experiments is heat loss to the surroundings, which can be mitigated by using better insulation or heat shields.
Q & A
What is calorimetry?
-Calorimetry is the process of determining the enthalpy of a reaction by measuring the heat absorbed or released during the reaction.
What is a calorimeter?
-A calorimeter is an instrument used to measure the enthalpy change of a reaction by observing the temperature change of a known amount of substance.
What are the typical components of a calorimeter setup?
-A typical calorimeter setup includes an insulated cup containing a known mass of water and a thermometer to measure the temperature change during a reaction.
How is the enthalpy change of a reaction calculated?
-The enthalpy change of a reaction is calculated using the formula ΔH = -q/n, where q is the heat absorbed or released by the system and n is the number of moles of product formed.
What is q in the context of calorimetry?
-In calorimetry, q represents the heat absorbed or released by the substance, which can be calculated using the formula q = mcΔT, where m is the mass of the substance, c is the specific heat capacity, and ΔT is the temperature change.
What is the specific heat capacity and why is it important in calorimetry?
-The specific heat capacity is the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius or one Kelvin. It is important in calorimetry because it is used to calculate the heat absorbed or released by the substance during a reaction.
What is the molar enthalpy of neutralization?
-The molar enthalpy of neutralization is the enthalpy change in a neutralization reaction when exactly one mole of water is formed.
Why is the molar enthalpy of neutralization typically negative?
-The molar enthalpy of neutralization is typically negative because neutralization reactions are exothermic, meaning they release heat.
What is the typical temperature and pressure for laboratory settings when measuring enthalpy changes?
-The typical temperature and pressure for laboratory settings when measuring enthalpy changes are 25 degrees Celsius and 100 kilopascals, respectively.
Why would a weak base be preferred over a strong base when neutralizing a chemical spill, such as sulfuric acid?
-A weak base would be preferred because it can absorb the liquid of the acid, limit the spread of the acid, and release less heat during the reaction, reducing the risk of additional dangers in a classroom setting.
What is the main source of error in a calorimetry experiment?
-The main source of error in a calorimetry experiment is heat loss to the surroundings, which can result in a lower measured q value than in theory.
How can the experiment be improved to prevent heat loss?
-The experiment can be improved by using double insulation, such as double-walled styrofoam cups, to minimize heat loss to the surroundings.
Outlines
🔍 Introduction to Calorimetry and Enthalpy Measurement
This paragraph introduces the concept of calorimetry, which is the science of measuring the enthalpy change of a reaction. It explains that a calorimeter is an instrument used for this purpose. The typical setup for calorimetry in an educational setting is described, involving an insulated cup with water and a thermometer to measure temperature changes due to chemical reactions. The goal is to measure the enthalpy change by observing the heat absorbed or released, changing the water's temperature. Key equations are introduced, such as enthalpy change being equal to the negative heat quantity (q), and the formula for heat (q = mcΔT), where m is mass, c is the specific heat capacity, and ΔT is the temperature change.
🧪 Understanding Calorimeter Setup and Neutralization Reactions
The paragraph delves into a typical calorimeter setup, including the use of a thermometer and an insulated cup filled with water to measure temperature changes during a reaction. It discusses the concept of molar enthalpy of neutralization, which is the enthalpy change when one mole of water is formed in a neutralization reaction. The standard conditions for such measurements are room temperature and pressure. The paragraph also touches on the practical considerations of choosing a weak base over a strong base for neutralizing chemical spills to control heat release and limit damage.
📚 Calorimetry Calculations and Experimental Procedure
This section focuses on the practical application of calorimetry calculations, starting with determining the number of moles of reactants using given concentrations and volumes. It emphasizes the importance of identifying the limiting reagent to understand how much of the reactants are consumed in the reaction. The goal is to find the number of moles of water produced, which is crucial for calculating the enthalpy change using the formula ΔH = -q/n. The paragraph also discusses the significance of accurate measurements and the impact of rounding on the final results.
🔢 Analyzing Calorimetry Data and Error Sources
The paragraph discusses the process of analyzing data from a calorimetry experiment, including calculating the heat absorbed by the water (q), the mass of the water, and the final temperature change. It highlights the importance of maintaining significant figures in scientific reporting and the potential discrepancies between theoretical and experimental values. The main source of error in such experiments is identified as heat loss to the surroundings, which can be mitigated by using double insulation or other heat shields.
🛠 Improving Calorimetry Experiments and Reflecting on Accuracy
In this paragraph, the discussion revolves around improving the accuracy of calorimetry experiments by addressing heat loss issues. Suggestions include using double insulation or reflective materials to minimize heat loss. The importance of using a sufficient amount of reagents to ensure a noticeable temperature change is emphasized, as it is crucial for accurate readings. The paragraph concludes with a reminder of the challenges in achieving precise measurements and the inherent limitations of experimental setups.
Mindmap
Keywords
💡Calorimetry
💡Enthalpy
💡Calorimeter
💡Temperature Change
💡Specific Heat Capacity (c)
💡Molar Enthalpy of Neutralization
💡Limiting Reagent
💡Exothermic Reaction
💡Heat Loss
💡Significant Figures
Highlights
Calorimetry is defined as the process of determining the enthalpy of a reaction.
A calorimeter is an instrument used to measure enthalpy changes during reactions.
Typical calorimetry setup includes a thermometer in an insulated cup with water or another substance to measure temperature changes.
The enthalpy change of a reaction can be calculated using the mass of water, specific heat capacity, and temperature change.
The formula for calculating the heat (q) absorbed or released by water is q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is temperature change.
The specific heat capacity (c) of water is 4.18 joules per gram per kelvin, a value important for accurate calorimetry calculations.
Molar enthalpy of neutralization is the energy change when one mole of water is formed in a neutralization reaction.
Standard conditions for laboratory settings are typically 25 degrees Celsius and 100 kilopascals.
The typical molar enthalpy of neutralization for strong acids and bases is around -57 kilojoules per mole.
Weak acids or bases result in a lower enthalpy change compared to strong acids or bases due to differences in bond energies.
In a chemical spill, a weak base like sodium carbonate is preferred over a strong base to control heat release and limit acid spread.
All neutralization reactions go to completion regardless of the strength of the acid or base involved.
The importance of rounding to the correct number of significant figures in chemistry calculations is emphasized for exam accuracy.
Heat loss to the surroundings is a primary source of error in calorimetry experiments.
Strategies to reduce heat loss include using double insulation and reflective materials like aluminum to retain heat within the system.
The significance of using a noticeable temperature change for accurate readings on a thermometer is discussed.
The transcript includes a step-by-step walkthrough of a calorimetry problem, illustrating the process of finding the limiting reagent and calculating the final temperature.
Transcripts
joe tell me about calorimetry
uh calorimetry is determining the
enthalpy of a reaction
good the goal is to measure
enthalpy triangle h reset tell me what a
calorimeter is
[Music]
yeah um
so a calorimeter is an instrument that's
used to measure enthalpy right that's
literally what it is it's the instrument
that we use in calorimetry
to actually measure the enthalpy
now gerald what was a typical setup for
the color emitter that you might have
done in year 11
um
so you would have a thermometer
sticking inside an insulated
cup and there would be water or some
other substance which you could measure
the temperature change of um during the
reaction good exactly right so the
perfect setup is right here
right so we have a volume of water that
is going to be absorbing any heat change
that occurs due to the reaction so
imagine we drop substance a and b
settles right to the center of the
solution and we have a chemical reaction
occurring
and that reaction is going to absorb
heat or it's going to release heat and
thus change the temperature of the water
does everyone agree
now we can measure the temperature
change using the thermometer that's
being stuck in right here
and we can use that
and the mass of the water to determine
the enthalpy change of reaction
okay so what are the two equations we
need to measure enthalpy change so
enthalpy change
equals to negative q on it you might
have seen that before
there is a form of a queue
that what is q equal to
um
q is equal to
seen mcat have you heard of q equals to
anything okay
yeah i've just looked back to that do it
so q refers to the energy
that is absorbed or released by the
water right so the energy
change
of
the body of water
so we need the mass of water does
everyone see that
we need the mass of water here it could
be 200 grams 100 grams etc so this is
the mass of water in grams
what is c gerald do you want to define
what c is
um that's the specific heat capacity of
the
substance
good what does that mean
um
it's the amount of energy required per
mole to change the temperature of that
substance by one kelvin
is it per mole
um
the unit is joules
per gram per kelvin
right so it's the amount of energy in
joules required to raise the temperature
of one gram of that substance by one
degree celsius or one degree kelvin so
it's joules
per gram times kelvin does that make
sense
and for water the value is 4.18 you
should know that it's in your data sheet
but i would commit it to memory to save
time
okay and finally t is a temperature
change
now
let's go up to the earlier reaction
now this reaction here is everything to
do the body of water i want you to
remember that q equals m cat is all
about the water don't run with me
whereas the n value here do you see this
n here
it has everything to do the actual
reaction that is going on inside of the
water
n is the number of moles of product you
produce
okay that's what n is
and so you can use this reaction to thus
find the energy change per mole of
substance produced
does that make sense that is how we
determine enthalpy changes
good now
who's learnt molar enthalpy of
neutralization at school
okay well we'll describe it so i'm
actually not gonna show you the
definition you're gonna intuitively
derive it
okay
molar enthalpy of neutralization but
what did we say enthalpy was gerald
that's the heat i mean g within the
system good right so if we're looking at
enthalpy of neutralization we're
measuring
the energy
of neutralization
and it's molar so what does that mean
it's per mole
and what is the one product you looked
at the net ionic equation of
neutralization what is the one product
that's always fall
water good so the molar enthalpy of
neutralization is the enthalpy change
in a neutralization reaction
when exactly one mole of water is formed
now
in general lab settings what would the
pressure and temperature
because you've got to standardize this
right you could get a different molar
enthalpy if you're doing it in the
antarctic or whether you're doing it in
you know uh uluru so
what's the typical temperature and
pressure in a
lap
25 degrees celsius and 100 kilopascals
exactly room temperature pressure
conditions so that is the literal
word-for-word definition have a look at
this the energy in kilograms per mole
liberated per mole of water
in rtp room temperature and pressure
now what did i say that number typically
was
when we did our net ionic equation
question it's typically negative
57
okay so typically entropy of
neutralization is negative 57 kilojoules
per mole
and that is because all acid-base
reactions have the same net ionic
equation now i want to stress this
this negative 57 happens when you have a
strong acid strong base
so if you have a strong acid weak base
you will liberate less energy than this
okay and that is to do with the bonds in
that weak base so remember if you ever
have a weak acid or weak base it's going
to lower this enthalpy change
everyone with me
so let me give you a thought experiment
this is how they test you in an exam
question if a student spilled
concentrated sulfuric acid
in the classroom
what would be an appropriate base
to neutralize that spill
and why
pick any base you have freedom of choice
would you pick a strong base like sodium
hydroxide or would you pick a weak base
like sodium carbonate
strong place
if you pick a strong base okay and what
about you gerald
oh weak base
okay
now
jared why would you pick a weak base
um because you ideally want to decrease
the um
actually
actually wait can i change my answer
are you sure you want to
explain your original thought you
wouldn't
originally it was um because you would
want to um probably decrease damage to
surroundings by decreasing um change in
l2p or the heat which is produced so you
would use the weak base you're
completely right actually right so you
need to understand this even with a
strong acid strong base or a weak acid
strong base so imagine imagine we had
hcl which is a strong acid and it
combines with a weak base like sodium
carbonate
the reaction will always go to
completion
you never form an equilibrium i want you
to remember this all neutralizations go
to completion
so we don't care about how strong the
acid or base is what we care about now
is the enthalpy i told you it's highly
exothermic so if we were to put sodium
hydroxide
on the hydrochloric acid
that would cause extreme amounts of heat
release in fact you could get boiling of
the solution
right and that would be a whole other
additional danger to the classroom
setting does that make sense the other
reason we actually prefer to use sodium
carbonate is because this comes as a
solid it looks like sand
right you've seen baking soda before
right the sodium carbonate has a similar
solid composition so it can actually
absorb the liquid of the acid as well
and thus limit the spread of the acid
physically
so that's why we prefer to use a weak
base that is a solid
over a strong base in chemical spills
this question has come up before so
that's why it's higher to understand
that
all right are we ready to start doing
some exam questions
all right so i'm going to throw you in
the deep end here is your
here's your first question
joe how about you uh i've got an answer
but um
i'm a bit suspect about the way i would
do that so yeah okay what's your answer
what's the final time
i got 25 points
25.75 degrees celsius
okay
okay
how many sig figs is in the question
though
um
three
is it three what's the lowest six you
see
um minus fifty seven point two
what about point one
potassium hydrogen oh right
make sure your answer is always to the
correct sig figs because there'll always
be one question in the hsc that they've
marked sig figs on and if you don't do
it right for that one question you'll
lose a mark
one mark is bigger than hsc one mark
drops you down thousands of
ranks and you're um you can go from a
state rank to a mid band six in one
single mark so that's why it's so
important
okay
let's start going through it so we'll
start with you gerald what's the first
thing i should do
um first thing i did was calculate the
number of moles of potassium
equation right
so what is the equation
uh koh plus hcl
and that gives you h2o plus kcl good
all right
balanced or not balanced all right
that's balance so okay so now tell me
what data do we have
um so
you have the
um volume of volume and concentration of
potassium hydroxide good i want you to
say c and v give it to me mathematically
so c of k koh is is equal to 0.1 molar
okay and what about v of koh and that is
equal to 27.9 milliliters
okay done
that's c and v what about uh
hcl um c of hcl is equal to 0.11 molar
okay and what's v of hcl and v of hcl is
equal to
31.25
milliliters
okay what else do we have
what are the data
and then we also have the initial
temperature
okay and let's say i'm stuck now what
should i do
um
as in like what's the next step yeah
uh so i calculated number of moles of
potassium hydroxide and hcl
good sure that works out what you could
also do is draw a diagram if you'd like
and then like i said frame the question
what is it actually asking it's asking
for tf
equals to question mark does that make
sense
so if we want to find tf we've got ti
we need to find change in t so the true
question is what is the change in
temperature if you find change in t
you've got your answer
right that's what i mean by framing the
question good so
number of moles of koh what did you get
so tell me for n of ko h n of hcl what
did you get
for nfk of h i got
0.00279 volts
and um for hcl i got
point zero zero three four three seven
five volts
joe do you get the same
uh yeah yeah that's right okay so why
did we find the number of moles of each
reason what are we trying to do
now uh we're trying to find which one is
the
uh limiting limiting reagent to find out
how much of it is are actually used up
good okay all right why don't we worry
about the limiting reagent gerald what's
the end goal of all this limiting
reagent work because we want to find out
number of moles of h2o produce very good
right because that's our n value if you
remember the equation triangle h
equals negative q and n well you've got
triangle h did you all see that so i'll
write that as a triangle h equals
negative 57.2 kilojoules per mole so
you've got triangle h if we can use
limiting axis to find n we can find q
and we've got all the other data so we
can thus find change in t
so the approach is actually quite
straightforward so now it's just about
following through so good so limiting
reagent would be
koh right
reset how do you find the number of
moles of water
uh use the stoichiometry so that you
know that then the number moles of
h2o is also the number of moles of kvh
which is
a let's call this as b so yeah that's
going to be equal to a moles
very good
so now we can use triangle h equals
negative q on n and we can find q
q equals to triangle h
times n or negative triangle h
so this is going to be negative of
negative 57.2 kilojoules remember it's
always exothermic per mole
times
a moles now do you see i'm writing units
here i don't usually do that but the
reason i'm doing that
is because triangle h is always in
kilojoules per ml whereas when you use q
you must convert it to joules per mole
okay that is why units is very important
so always be cautious of that it's a
huge silly mistake students make going
from joules to kilojoules so
what's our answer for q
tell me the answer in kilojoules and
joules gerald
um so in kilojoules i got minus 0.159588
so minus zero point
one five
nine five
eight eight
okay let's do that as should shouldn't
it be plus here because it should be
it's minus or minus so it should be plus
remember q is the energy absorbed by the
water if it's an exothermic reaction
think about it where did all that heat
go to the water from the system so it
has to be plus and mathematically it
makes sense because it's an exothermic
reaction
and q is negative
negative of an exothermic or negative
value is positive does that make sense
good so that's our kilojoules but let's
convert it to joules so that's going to
be is it
159.58 joules
uh yeah
good let's store that as d all right
okay so now we know q equals to m cap
what's our mass value
what's a mass of water
rheostat
is it
27.9 plus 31.25 which is exactly right
it's all just water that's what the
solution is so what is that
59.15 millions
good
so convert that to liters and what's
this is the heat capacity of water 4.18
so what's our final answer
um
so
okay
okay
uh yeah
what did we get
i got 25.6 degrees celsius i mean 0.6
that's the change
okay what was the full number
what was the full number was it exactly
0.6 what's the full number oh full
number was
well it was long it was like 0.6454
[Music]
yeah okay good so round that off to x
good and the final thing we can say is
well
well we know that change in t is equal
to t f minus t i we're trying to find tf
so tf
is change in t plus ti so the answer is
equals to 25.645
you said 25.7 is that because you
rounded it but it wouldn't round up
oh
um
i think it's six four five i think the
last time i'd done this question it was
point six four five double check you're
working yeah yeah i think i might have
rounded a bit too early somewhere yeah
never around in chemistry until the very
end you want to make this
practice never ever always store it and
at the very end you then round it
okay because they can mark you down even
if you're off by the decimal points so
good so then this will round off
to final temperature being 30 degrees
because it's to one sig fig
um if i still got 30 degrees well like
yeah if i still got 30 degrees celsius
like the um to signal never configures
even if my um
like
unrounded answer was like slightly off
would that be a problem
it really depends on your marker see
that's the thing you're i think we have
to realize that schools are very
non-systematic they do whatever they
like right so what would happen in that
case is
if you set your exam and you did that
and if there was a very small standard
deviation in the cohort then they would
make that part of the criteria and you
would lose the mark okay so that's
typically what happens
in the hsc
they try to be lenient so they may give
it to you but certain questions it's
very strict it all depends what was made
as a marketing criteria
so i would say to be safe you want to
have a perfect answer rendered off at
the very end okay
good
all right good job homework is going to
be
you should have access to this booklet i
will double check your access if you
don't have access throughout mid next
week do message me okay and i will uh
i'll follow through on that for us
so
page 20
21
22 are all additional practice questions
of the same type so you can do that in
your own time okay
if you can do those questions it's very
easy and i guess the one thing we'll
talk about
discrepancy between theoretical and lab
values so reset what would the main
discrepancy be here
what would result in a discrepancy
between the true value and our
calculated value for enthalpy change
what is the main source of error in this
experiment
come on there's a reason we use double
double styrofoam cups
uh energy released into the surroundings
exactly right and that results in a
lower q value
than in theory
right because
yeah sorry my parents are here and they
just need me to open them so i'll just
okay no worries okay so um yeah so
that's what i'm mentioning the main
issue here is heat loss to the
surroundings
heat loss okay gerald how would you
improve this experiment to prevent the
heat loss
double insulated so what else yeah
um
use less
reactants such that the change in
temperature isn't great such that so you
won't lose as much heat to the
environment
uh change in temperature is not so great
that you will not
but see the the when we talk about
accuracy it's about percentage error
right so the percentage area will still
be there even if the temperature change
is larger i get where you're coming from
um
i would say if you used more reagent you
cause more heat to change
i think that's good because if you think
about this if you use a very tiny amount
of reagent what if your thermometer
can't even pick up the temperature
change accurately that's the limiting
factor right if the temperature change
is 0.0002 degrees celsius your
thermometer won't even pick up a
temperature change and thus you'll be
limited by the accuracy of your
instruments so i would say you'd want a
noticeable temperature change so you can
actually pick it up accurately on a
thermometer
but good thought good thought but um i
would say one of the main things is heat
shield so you could use
certain structures like aluminium and
that can refract reflect heat back into
the system
but uh again it's a bit wishy-washy it
doesn't work too well
good let's move on
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