# LAB!!

posted by .

I have to find the molar enthalpy of combustion of pentane. I have this info:

5C(s) + 6H2 >>> C5H12 Hf= -173.5kJ
C+O2>>> CO2 Hf= -393.5kJ
H2 + 1/2O2 >>> H2O Hf= -241.8kJ
H2O >>> H2O Hf= +44.0kJ

THEN I am given this:
mass of pentane= 2.15g
volume of water equivalent to calorimeter= 1.24L
initial temp. of calorimeter & contents= 18.4 degrees celcius
final temp. of calorimeter & contents= 37.6 degrees celcius

The one thing I don't get is why the give me the equations with the Hf. I don't get it. As well, what am I supposed to do with the rest of the information? Can you please help me get started. Thanks*

The equations, I think, are there to allow you to calculate, from the values given, the heat of combustion that you SHOULD get from experimental data. Reverse equation 1, then add equation 2 (multiplied by 5) and add equation 3 (multiplied by 6). You left something off the last equation but I assume it is H2O liquid ==> H2O steam (or it may be the other way around). You will need to add or subtract that 44 kJ (remember that is per mole and you have 6 mols H2O ) as appropriate. Also, remember that if you multiply an equation by 5 or 6 to also multiply the kJ and if you reverse the equation you must reverse the sign given. Finally, the last part is an experimental part.
Delta H = mass x specific heat water x (Tf - Ti) where Tf is the final temperature and Ti is the initial temperature. I would assume the density of water to be 1.00 g/L here so the mass of H2O would be 1,240 g. The specific heat of water is 4.18 J/g*C. This should give you the experimental value for the combustion of pentane FOR 2.15 GRAMS. You want the value per mol. I hope this helps.

• LAB!! branch -

i need help

## Similar Questions

1. ### kim

a gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. given the following data, determine (delta)HoF for hydrocarbon: (delta)Horxn=-2044.5kj/mol hydrocarbon (delta)HoF CO2= -393.5kj/mol (delta)HoF …
2. ### Dr. Bob

a gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. given the following data, determine (delta)HoF for hydrocarbon: (delta)Horxn=-2044.5kj/mol hydrocarbon (delta)HoF CO2= -393.5kj/mol (delta)HoF …
3. ### Chemistry

Data: C(graphite) + O2(g) forms CO2 (ga) Delta H = -393.5kJ H2(g) + 1/2O2(g) forms H2O(l)Delta H = 285.8kJ. CH3OH(l) + 3/2O2(g)forms CO2(g) + 2H2O(l)Delta H = -726.4 Using data above, calculate the enthalpy change for the raction below …
4. ### Chemistry

Caluclate the delta H for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of liquid pentane is -146.8kJ/mol. Balanced equation balanced equation: C5H12(l)+8O2(g)-->5CO2+6H2O(g) CO2=393.5 …
5. ### chem

please help me? :) topic: standard enthalpies of formation "Calculate ΔHf° of octane, C8H18(l), given the entalpy of combustion of octane to CO2(g) and H2O(l) is -5471kJ/mol. The standard enthalpies of formation ofCO2 and H2O
6. ### chemistry

molar enthalpy of combustion for pentane?
7. ### Chemistry

I am having trouble with this equation and calculating the enthalpy of formation. 2 NaHCO3 (s) --> Na2CO3 (s) + H2O (g) + CO2 (g) delta H129.2kj I get as far as this step and then I don’t know what to do next. delta H= [1*delta …
8. ### Chemistry

Calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6, from the following data: 2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l) ΔH∘ = -6534kJ ΔH∘f (CO2) = -393.5kJ/mol ΔH∘f (H2O) = - 285.8kJ/mol
9. ### Chemistry

We have of ice at -11.9°C and it is ultimately completely converted to liquid at 9.9°C. For ice, use a specific heat (Cs) of 2.01 J/gK, for liquid water use a specific heat of 4.18 J/gK, and ΔHfusion = 6.01kJ/mole, Molar mass …
10. ### Chemistry

Determine the Enthalpy of combustion of ethyne (C2H2) using the following enthalpies of formation; Data: DHf°(H2O)= -285.9kj/mol DHf°(2H2)= +226.7Kj/mol DHf°(CO2)= -393.5Kj/mol Equation; C2H2--------->CO2 + H2O

More Similar Questions

Post a New Question