Question

A calorimeter contains 500 g of water at 25°C. You place a hand warmer containing 100 g of liquid sodium acetate (NaAC) inside the calorimeter. When the sodium acetate finishes crystallizing, the temperature of the water inside the calorimeter is 32.2°C. The specific heat of water is
4.18 J/g-°C. What is the enthalpy of fusion (ΔHf) of the sodium acetate? Show your work.

Answer 1

So, I know that enthalpy of fusion is q=mC∆T (right?). But from there I'm kind of confused. Do I use 500 g as m or 100 g as m? And do I just solve from there or are there other things I have to do to get the correct answer? Thanks!

Answer 2

The sodium acetate is the substance producing heat, the water is ABSORBING the heat and in turn changing temperature. SO the temperature of the water is what were interested in because it's measuring the heat released from the crystallization.

I'm not sure if the question is asking you to find the answer per mole of sodium acetate, (because it's not specifically stating "standard enthalpy of fusion") but if it is, use:

\(\Delta H^o_{fusion}=\underbrace{\Delta H_{fusion}}_{found~above}*n_{sodium~acetate}\)

Answer 3

By the way, this is after you've used: \(q=m*C*\Delta T\) to be clear.

Answer 4

So q = mC∆T would be
q = (100 g)(4.18 J/g-°C)(32.2 °C - 25°C)
q = (100 g)(4.18 J/g-°C)(7.2°C)
q = 3009.6 J
correct?

Answer 5

you need to use the water because that's the medium you're measuring the temperature change through.

Answer 6

Okay so it's
q = (500 g)(4.18 J/g-°C)(32.2°C - 25°C)
q = (500 g)(4.18 J/g-°C)(7.2°C)
q= 15048?

Answer 7

correct. remember that were measuring the heat given off by the crystallization, so the answer at the end MUST be negative.

It's positive here because we're measuring the heat absorbed but the water (which is positive)

Answer 8

absorbed by the water**

Answer 9

Okay! I'm looking back in my notes, so just to make sure I understand, the ΔH we have right now is positive because the reaction is endothermic (the water is ABSORBING the heat)? Just wanna check my understanding, this unit is kinda tricky for me

Answer 10

yes, thats exactly it. This is why we get a positive value. But when you report the \(\Delta H_{fusion}\) of sodium acetate it has to be negative.

Answer 11

So to get the ΔHfusion of sodium acetate we use the formula you gave above (ΔHf x # of moles of sodium acetate) or...

Answer 12

yeah, it depends. If they're asking for the \(\Delta H_{fusion}\) for the reaction, then you've already found it.

But if they're asking for the standard enthalpy, which is denoted by the circle on \(\Delta H^o_{fusion}\) (and in units of energy unit/mol), then you need to use the formula above.

I'm not sure what they want, but I'm leaning more towards the standard because they gave you the mass of NaAcO.

Answer 13

Your guess is as good as mine. I looked into lessons after this assignment and they do talk about using kJ/mol, but I asked myself why would they ask us to solve a problem when we don't get into that yet? But then I remembered I take Honors Chemistry so we usually get challenged like this, so you're probably right ^^

Answer 14

yeah if the units are in "kJ/mol", then you definitely need to use he standard. Good eye.

Answer 15

Question, I was calculating the molar mass of NaAc, Na: 22.99 and Ac: 227 so the molar mass should be 249.99 g/mol right? But when I googled the molar mass of NaAC it came up as 82 g/mol?

Answer 16

Ac is short for acetate \(C_2H_3O_2^-\), you wrote the molar mass of Ac which is the element actinium.

Answer 17

the molar mass should be (82 +22.99 )g/mol

Answer 18

Ohh okay. So molar mass is 82 g/mol.
# of moles of NaAc: 100 g/82 g/mol= 1.2 mol
15048 J = 15.048 kJ
ΔH°f = 15.048 J x 1.2 mol = 18.0576 kJ/mol? Is that right?
Or do we divide the ΔHf by the number of moles to get kJ over mol.

Answer 19

damn, i wrote it wrong. you do have to divide

\(\Delta H_{fusion}=\Delta H^o_{fusion}*n\rightarrow \Delta H^o_{fusion}=\dfrac{\Delta H_{fusion}}{n}\)

Answer 20

That's fine! Okay, so its (15.048 kJ)/(1.2 mol) = 12.54 kJ/mol?

Answer 21

yes, thats it !

Answer 22

Woo! So that's the final answer right? I hope so, I'm so tired of working on this problem, haha

Answer 23

haha yes that's the final answer, just remember that it's a negative value.

Answer 24

I saw your response earlier, but I was out and for some reason I couldn't reply on my phone. But yeah, thanks so much for your help! You're a lifesaver :D

Answer 25

no problem ! glad i could help

Answer 26

was this the correct answer?