# Thermochemistry and the heat of complete combustion

• Chemistry
• DottZakapa
In summary, the balanced reaction is 2C6H6(l) + 15O2(g) -> 12CO2(g) + 6H2O(l). To compute the standard enthalpy of reaction, we use the given standard enthalpies of formation for H2O(l), CO2(g), and C6H6(l) to calculate the heat of reaction. Converting 4,00L to 4000mL, we can determine the grams of C6H6(l) formed to be 3120g. By proportion, the moles of liquid water formed will be 120 mol. Using this information, we can calculate the heat per mole of benzene and the total heat

#### DottZakapa

Homework Statement
Calculate the heat Kcal measured at 25.0°C and 1 atm corresponding to the complete combustion of 4.00 L of Benzene (C6H6 d= 0.780 6/mL) when liquid water forms.
Relevant Equations
thermochemistry
The balanced reaction wil be :
##2C_6H_{6(l)}+15O_{2(g)}->12CO_{2(g)}+6H_2O_(l)##
in order to compute the the standard enthalpy of reaction :
##\Delta H°_{f} H_2O_(l)= -285,8 \frac {KJ}{mol}##;
##\Delta H°_{f} CO_{2(g)}= -393,5 \frac {KJ}{mol}##;
##\Delta H°_{f} C_6H_{6(l)}=49,04 \frac {KJ}{mol}##;

##\Delta H°_{R}= \sum \Delta H°_{f} \text{ products} -\sum \Delta H°_{f} \text{ reagents}= [(-285,8 \frac {KJ}{mol}*6\text{ mol})+(-393,5 \frac {KJ}{mol}*12\text{ mol})]-(49,04\frac {KJ}{mol}*2\text{ mol})=-6531,26 KJ##

converting L to mL:
4,00L=4000 mL
The grams of ##2C_6H_{6(l)}##:
4000 mL *0,789 ##\frac g {mol}##= 3120 g;

Moles of ##2C_6H_{6(l)}##:
##\frac {3120 g}{6*12,01\frac g {mol}+6*1,01\frac g {mol} }##= 39,9 mol;

By proportion the moles of liquid water formed will be :
mol ##H_2O_(l)##=## \frac{6 \text{ mol } H_2O_(l)*39,9\text{ mol } 2C_6H_{6(l)}}{2 \text{ mol }2C_6H_{6(l)}}##= 120 mol

The heat , by proportionality :
##\frac{120 \text{ mol } *(-6531,26 KJ)}{6 \text{ mol }}##= -131*10^3 KJ = -3,13*10^4 Kcal

Is it all correct in your opinion?
the book says 3,03 *10 ^4 Kcal
the allowed error is 3% by the professor, where am i wrong?

The amount of heat that has to be removed per mole of benzene to hold the temperature constant at 25 C is 6531.26/2 = 3265.63 Kj. For 39.9 moles burned, this comes to 130300 Kj.

Last edited:
thanks,ok now i got why is positive, but now still i don't get why the result doesn't match the one on the book 3,03 *10 ^4 Kcal, even yours, divided by 4,184 still not matching

DottZakapa said:
thanks,ok now i got why is positive, but now still i don't get why the result doesn't match the one on the book 3,03 *10 ^4 Kcal, even yours, divided by 4,184 still not matching
It's not worth worrying about.

well, as i said, at the exam there is an error tolerance of 3%, but this still on the range more or less. Thanks anyway

## 1. What is thermochemistry?

Thermochemistry is the branch of chemistry that deals with the study of energy changes in chemical reactions and the relationship between heat and other forms of energy.

## 2. What is the heat of complete combustion?

The heat of complete combustion is the amount of heat released when one mole of a substance is completely burned in excess oxygen under standard conditions.

## 3. How is the heat of complete combustion calculated?

The heat of complete combustion can be calculated by using the enthalpies of formation of the products and reactants in the balanced chemical equation and applying the Hess's Law.

## 4. Why is the heat of complete combustion important?

The heat of complete combustion is important because it provides information about the energy content of a substance and its potential as a fuel source. It also helps in predicting the amount of heat released or absorbed in a chemical reaction.

## 5. How does the heat of complete combustion affect the environment?

The heat of complete combustion can have a significant impact on the environment, as it is directly related to the amount of carbon dioxide and other greenhouse gases released into the atmosphere. The higher the heat of combustion, the more pollutants are released, contributing to global warming and air pollution.