The enthalpy of atomisation is the enthalpy change upon breaking the bonds of 1 mole of a gaseous compound into its atom constituents. Using the table of average bond enthalpies below calculate the enthalpy of atomisation of ethene (C2H4): Bond Average bond enthalpy /kJ mol-1
C=C 612
C-H 412
The enthalpy of atomisation of ethene is:
Can someone help me as this is really confusing me. Thanks in advance
ethene has one C=C bond , and four C-H bonds
(1 * 612) + (4 * 412) = ?
Sure, I'd be happy to help you with the calculation! But first, let me lighten the mood with a little joke:
Why did the scarecrow win an award?
Because he was outstanding in his field!
Alright, let's get down to business. To calculate the enthalpy of atomisation of ethene (C2H4), we need to break all the bonds in 1 mole of ethene and determine the change in enthalpy.
Let's start by breaking the carbon-carbon double bond (C=C). The bond enthalpy for a C=C bond is 612 kJ/mol. Since ethene has two C=C bonds, the enthalpy change for breaking them both would be 2 * 612 kJ/mol.
Next, we need to break the carbon-hydrogen (C-H) bonds in ethene. Ethene has four C-H bonds. The bond enthalpy for a C-H bond is 412 kJ/mol, so the enthalpy change for breaking the four C-H bonds would be 4 * 412 kJ/mol.
Now, let's sum up all the bond dissociation energies:
Enthalpy change for breaking C=C bonds: 2 * 612 kJ/mol = 1224 kJ/mol
Enthalpy change for breaking C-H bonds: 4 * 412 kJ/mol = 1648 kJ/mol
Finally, to calculate the enthalpy of atomisation, we add up the two enthalpy changes:
Enthalpy of atomisation of ethene = Enthalpy change for breaking C=C bonds + Enthalpy change for breaking C-H bonds
= 1224 kJ/mol + 1648 kJ/mol
= 2872 kJ/mol
So, the enthalpy of atomisation of ethene is 2872 kJ/mol. Keep in mind that this is an approximate value since bond enthalpies can vary depending on the molecule's environment and other factors.
I hope that clears things up for you! If you have any more questions, feel free to ask.
To calculate the enthalpy of atomisation of ethene (C2H4), we need to break all the bonds in the compound and then calculate the overall energy change.
Ethene (C2H4) has multiple bonds, including one C=C bond and four C-H bonds. We will consider breaking these bonds one at a time.
1. First, let's break the C=C bond:
Since the bond enthalpy for a C=C bond is 612 kJ/mol, breaking one C=C bond requires an input of 612 kJ/mol.
2. Next, let's break the four C-H bonds:
Since the bond enthalpy for a C-H bond is 412 kJ/mol, breaking one C-H bond requires an input of 412 kJ/mol. Since we have four C-H bonds in ethene, breaking all of them requires an input of 4 * 412 = 1648 kJ/mol.
3. Now, let's calculate the total energy change for breaking all the bonds:
Total energy required = energy to break C=C bond + energy to break C-H bonds
= 612 kJ/mol + 1648 kJ/mol
= 2260 kJ/mol
Therefore, the enthalpy of atomisation of ethene (C2H4) is 2260 kJ/mol.
To calculate the enthalpy of atomization of ethene (C2H4), we need to consider the bonds that need to be broken to break down the molecule into its individual atoms.
Ethene (C2H4) consists of two carbon (C) atoms and four hydrogen (H) atoms. We need to know the bond enthalpies for the C-C and C-H bonds to calculate the enthalpy of atomization.
The enthalpy change for the breaking of bonds is always endothermic (positive value) because energy is required to break the bonds. The enthalpy of atomization is the sum of the bond enthalpies for all the bonds broken.
For ethene (C2H4), the bonds that need to be broken are:
- 1 C-C bond
- 4 C-H bonds
Using the average bond enthalpies given in the table, we can calculate the enthalpy of atomization.
Enthalpy of atomization = (Number of C-C bonds x Bond enthalpy for C-C bond) + (Number of C-H bonds x Bond enthalpy for C-H bond)
Enthalpy of atomization = (1 x 612 kJ/mol) + (4 x 412 kJ/mol)
Enthalpy of atomization = 612 kJ/mol + 1648 kJ/mol
Enthalpy of atomization = 2260 kJ/mol
Therefore, the enthalpy of atomization of ethene (C2H4) is 2260 kJ/mol.