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.

i need help

To find the molar enthalpy of combustion of pentane, you can follow these steps:

1. Reverse equation 1 (5C(s) + 6H2 >>> C5H12) and write it as C5H12 >>> 5C(s) + 6H2.

2. Multiply equation 2 (C + O2 >>> CO2) by 5 to match the number of carbon atoms in pentane. The equation becomes 5C + 5O2 >>> 5CO2.

3. Multiply equation 3 (H2 + 1/2O2 >>> H2O) by 6 to match the number of hydrogen atoms in pentane. The equation becomes 6H2 + 3O2 >>> 6H2O.

4. Add the modified equations from steps 1, 2, and 3 together:

C5H12 + 5O2 >>> 5CO2 + 6H2O.

5. Add the enthalpy values for the reactants and subtract the enthalpy values for the products:

Enthalpy of combustion = [5(-393.5 kJ) + 6(-241.8 kJ)] - (-173.5 kJ) - (6 × 44.0 kJ).

Simplifying this equation will give you the enthalpy of combustion for 1 mole of pentane.

6. The given mass of pentane is 2.15 g, so you need to convert this to moles. The molar mass of pentane (C5H12) is 72.15 g/mol (5 × 12.01 + 12 × 1.01).

Moles of pentane = mass of pentane / molar mass of pentane.

7. Divide the enthalpy of combustion (step 5) by the moles of pentane (step 6) to get the molar enthalpy of combustion of pentane.

Now, let's move on to the experimental part:

8. Calculate the mass of water equivalent to the calorimeter using its volume and assuming a density of 1.00 g/mL.

Mass of water equivalent = volume of water equivalent x density of water.

9. Calculate the change in temperature (ΔT) by subtracting the initial temperature from the final temperature.

ΔT = final temperature - initial temperature.

10. Use the formula to calculate the experimental value for the combustion of pentane for 2.15 grams.

ΔH = mass x specific heat water x ΔT.

ΔH represents the heat gained or lost, the mass is the mass of water equivalent, and the specific heat of water is 4.18 J/g°C.

11. Convert the experimental value to per mole by dividing it by the moles of pentane (step 6).

These steps should guide you in calculating the molar enthalpy of combustion of pentane and obtaining the experimental value based on the given information.

To find the molar enthalpy of combustion of pentane, you can use Hess's Law and the given equations with their enthalpy values.

First, reverse equation 1 to get the combustion of pentane:

C5H12 + 8O2 >>> 5CO2 + 6H2O (Note: We have reversed the equation and changed the sign of the enthalpy value.)

Next, multiply equation 2 by 5 and equation 3 by 6 to balance the number of carbon and hydrogen atoms:

5C + 5O2 >>> 5CO2 (-5 * -393.5 kJ/mol = 1967.5 kJ/mol)
6H2 + 3O2 >>> 6H2O (-6 * -241.8 kJ/mol = 1450.8 kJ/mol)

Now, add the three equations together:

C5H12 + 8O2 + 5C + 5O2 + 6H2 + 3O2 >>> 5CO2 + 6H2O + 5CO2 + 6H2O
C5H12 + 16O2 + 5C + 5O2 + 6H2 + 3O2 >>> 10CO2 + 12H2O

Simplifying, we get:

C5H12 + 21O2 >>> 10CO2 + 12H2O (Note: The enthalpy change for this reaction is the sum of the enthalpies of the individual reactions.)

Now, sum up the enthalpy changes:

ΔH = -173.5 kJ/mol + 1967.5 kJ/mol + 1450.8 kJ/mol + 44.0 kJ/mol

ΔH = 2289.8 kJ/mol

This is the enthalpy change for the combustion of pentane per mole.

To find the enthalpy change for 2.15 grams of pentane, you will need to use the experimental data given:

mass of pentane = 2.15 g
volume of water equivalent to calorimeter = 1.24 L
initial temperature of calorimeter & contents = 18.4 degrees Celsius
final temperature of calorimeter & contents = 37.6 degrees Celsius

Use the equation:

ΔH = mass x specific heat of water x (Tf - Ti)

Here, the mass of water is equal to the volume of water (1.24 L) since the density of water is assumed to be 1.00 g/L. The specific heat of water is 4.18 J/g°C.

Plug in the values:

ΔH = 1.24 L x 4.18 J/g°C x (37.6°C - 18.4°C)
ΔH = 58.34 J

Convert J to kJ by dividing by 1000:

ΔH = 0.05834 kJ

This is the experimental value for the combustion of 2.15 grams of pentane.

To find the molar enthalpy of combustion, divide the experimental value by the number of moles of pentane:

moles of pentane = mass of pentane / molar mass of pentane

The molar mass of pentane (C5H12) is 72.15 g/mol.

moles of pentane = 2.15 g / 72.15 g/mol
moles of pentane = 0.0298 mol

Now divide the experimental value by the number of moles:

molar enthalpy of combustion = ΔH / moles of pentane
molar enthalpy of combustion = 0.05834 kJ / 0.0298 mol
molar enthalpy of combustion = 1.959 kJ/mol

Therefore, the molar enthalpy of combustion of pentane is approximately 1.959 kJ/mol.