Question

A 1.20-g sample of propane (C3H8) is burned in a bomb calorimeter. The temperature of the calorimeter rises from 20.0°C to 30.0°C. The calorimeter has a mass of 2.000 kg and a specific heat of 2.95 J/(g • °C). If the molar mass of propane is 44.0 g/mol, what is the molar heat of combustion of propane?

6,490 kJ/mol
4,330 kJ/mol
2,160 kJ/mol
59.0 kJ

Answer 1

First find the thermal energy absorbed by the water:

\(\sf q_{absorbed}=m_{water}*C_p*\Delta T\)

then scale it to "per mole" with the mass given to get the molar heat of combustion \(\sf H^o_{comb}\).

\(\sf q_{absorbed}=-q_{released}=- H^o_{comb}*n_{propane}\) (n=moles)

Answer 2

I have no clue where to start. I don't even see water in the question.

Answer 3

I have 11 Minutes to find out how to do this to get the answer.

Answer 4

water is used in the calorimeter to absorb the energy and increase in temperature. That's how we measure the energy released by the reaction

Answer 5

so there is 2kg of water.

Answer 6

They have integrated the mass of the water with the calorimeter.

The first equation should be: \(q_{abs}=m_{cal}*C^{cal}_p*\Delta T\)

Answer 7

But i don't understand how i would take the numbers and plug them into the equation.

Answer 8

m=mass, \(C_P\)=specific heat capacity, \(\Delta T=T_f-T_i\) (temp change)

Answer 9

10 C is ΔT

Answer 10

yep

Answer 11

2.95 is Cp

Answer 12

mass would be 1.20 g?

Answer 13

Would it be the mass of the sample or water.

Answer 14

nope, thats the mass of the propane (which is releasing the heat), you would use the mass ABSORBING the heat, which is the mass of the calorimeter, 2kg.

Answer 15

look at the units though, heat capacity is given in grams, so you need to convert

Answer 16

111Mol?

Answer 17

you're using mass here, not moles

Answer 18

Oh! then it would be 2000 grams

Answer 19

yep!, now multiply those values together to get q

Answer 20

59000

Answer 21

dont forget the units.

Now we wanna use the other equation to scale the heat released to a molar value.

We need the moles of propane, use \(\sf moles=\dfrac{mass}{Molar~mass}\)

Answer 22

so 59000/44

Answer 23

= 1337

Answer 24

not quite. You didnt find the moles of propane, you just divided the energy by the molar mass.

\(\sf H^o_{comb}=\dfrac{q}{\dfrac{mass}{Molarmass}}=\dfrac{q*Molar~ mass}{mass}\)

Answer 25

.027

Answer 26

Mol of propane

Answer 27

okay, now divide the heat released by the moles

Answer 28

so 59000 divided by .027

Answer 29

yep

Answer 30

you should always write units

Answer 31

Yeah, sorry, i got a hugggggggeeeeee number.

Answer 32

it should be pretty large

Answer 33

2,185,185.185

Answer 34

right, now convert it to kJ, 1000 J = 1 kJ

Answer 35

Then divide by 1000

Answer 36

yes

Answer 37

so it would be 2185.2 which is closest to C.

Answer 38

Kj/m

Answer 39

yep. remember to write mole, as "mol" not just "m"

Answer 40

Oops yeah, thank you!

Answer 41

no problem!