{Thermochemistry} Hess' law problem

[hiuting]

Homework Statement

Benzene (C₆H₆) is a liquid which burns in air to form liquid water and carbon dioxide. Calculate the heat released by burning one gram of benzene.

$$\Delta$$H^o_f, kJ/mol
carbon dioxide (g) = -393.5
water (l) = -285.8
benzene (l) = 49.04

Homework Equations

2C₆H₆ (l) + O₂ (g) >> 12 CO₂ (g) + 6 H₂O (l)

The Attempt at a Solution

I got -25.42 kJ ... do I have to find the amount of moles in 1 gram of benzene first? and then do the calculations?

 

[nate808]

Thank you for your question. To calculate the heat released by burning one gram of benzene, we first need to determine the amount of moles in one gram of benzene. This can be done by using the molar mass of benzene, which is 78.11 g/mol. So, one gram of benzene contains 0.0128 moles (1 g / 78.11 g/mol = 0.0128 mol).

Next, we can use the balanced chemical equation you provided to calculate the heat released. According to the equation, 2 moles of benzene react to form 12 moles of carbon dioxide and 6 moles of water. This means that for every 2 moles of benzene, 12 moles of carbon dioxide are produced. Therefore, for 0.0128 moles of benzene, (0.0128 mol / 2 mol) x 12 mol = 0.077 mol of carbon dioxide is produced.

We can now use the given values for the standard enthalpies of formation (\DeltaHof) to calculate the heat released. Using the equation \DeltaH = \sum n \DeltaHof(products) - \sum n \DeltaHof(reactants), we get:

\DeltaH = (0.077 mol)(-393.5 kJ/mol) + (0.0128 mol)(-285.8 kJ/mol) - (0.0128 mol)(49.04 kJ/mol)

= -30.34 kJ/mol

Since we are dealing with 0.0128 moles of benzene, we need to multiply this value by 0.0128 to get the heat released for one gram of benzene.

Therefore, the heat released by burning one gram of benzene is -30.34 kJ/mol x 0.0128 mol = -0.388 kJ.

I hope this helps answer your question. If you have any further inquiries, please don't hesitate to ask.