In `C_(2)H_(4)` energies of formation of `(C=C) and (C-C)` are -145 kJ/mol and -80kJ/mol respectively. What is the enthalpy change w hen ethylene polymerises to form polythene?

Options

A. `+650" kJ/mole"`
B. `1.5 " kJ/mole"`
C. `-650 " kJ/mole"`
D. `-65" kJ/mole"`

Answer

To find the enthalpy change when ethylene (C₂H₄) polymerizes to form polyethylene, we can follow these steps:

### Step 1: Identify the Bonds Involved
When ethylene (C₂H₄) polymerizes, the double bond (C=C) in ethylene breaks and forms single bonds (C-C) in the polymer (polyethylene).

### Step 2: Write the Reaction
The polymerization of ethylene can be represented as:
\[ n \, \text{C}_2\text{H}_4 \rightarrow \text{(C}_2\text{H}_4\text{)}_n \]

### Step 3: Determine the Enthalpy Change Formula
The enthalpy change (ΔH) for the reaction can be calculated using the formula:
\[ \Delta H = \text{(Enthalpy of products)} - \text{(Enthalpy of reactants)} \]

### Step 4: Calculate the Enthalpy of Products
In the polymer (polyethylene), the bonds formed are C-C single bonds. The energy of formation for a C-C bond is given as -80 kJ/mol.

### Step 5: Calculate the Enthalpy of Reactants
In the reactants (ethylene), the bonds present are C=C double bonds. The energy of formation for a C=C bond is given as -145 kJ/mol.

### Step 6: Substitute Values into the Formula
Now, we can substitute the values into the formula:
\[ \Delta H = \text{(Enthalpy of C-C)} - \text{(Enthalpy of C=C)} \]
\[ \Delta H = (-80 \, \text{kJ/mol}) - (-145 \, \text{kJ/mol}) \]
\[ \Delta H = -80 + 145 \, \text{kJ/mol} \]
\[ \Delta H = 65 \, \text{kJ/mol} \]

### Step 7: Conclusion
The enthalpy change when ethylene polymerizes to form polyethylene is \( +65 \, \text{kJ/mol} \).

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