Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The formation reaction of
will be,

The intermediate balanced chemical reaction will be,
(1)

(2)

(3)

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :
(1)

(2)

(3)

The expression for enthalpy of formation of
will be,



Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ
And what should I do with this information
✡ Answer: 1.23*10^2 ✡
- - Add a decimal at the end (to the right) and count till you get to the first number.
So now you have 1.23
- - Now you always want to times it by 10 to the power of how many times you moved it over, in this case, 2
Final answer: 1.23*10^2
✡Hope this helps✡
How many hydrogen atoms are involved in this reaction? 3
Elements in the same group have similar properties