The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):
![\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol](https://tex.z-dn.net/?f=%20%5CDelta%20H%20%3D%20%5B2%2A%286%2A413%20%2B%20347%29%20%2B%207%2A498%20-%20%284%2A2%2A799%20%2B%206%2A2%2A467%29%5D%20kJ%2Fmol%20)
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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Answer:
a. because it is an element
Explanation:
its pure because it only has one type of atom, making it an element
Answer:
55.3 × 10²³ molecules
Explanation:
Given data:
Number of moles of C₁₁H₁₂O₂₂ = 9.18 mol
Number of molecules = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
For given data:
9.18 mol × 6.022 × 10²³ molecules /1 mol
55.3 × 10²³ molecules
<h3>Question:</h3>
•Why do you feel the force of Earth's gravity?
Answer:
•Earth's gravity comes from all its mass. All its mass makes a combined gravitational pull on all the mass in your body. That's what gives you weight. And if you were on a planet with less mass than Earth, you would weigh less than you do here.
Explanation:
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