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
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Answer:
C) In[reactant] vs. time
Explanation:
For a first order reaction the integrated rate law equation is:

where A(0) = initial concentration of the reactant
A = concentration after time 't'
k = rate constant
Taking ln on both sides gives:
![ln[A] = ln[A]_{0}-kt](https://tex.z-dn.net/?f=ln%5BA%5D%20%3D%20ln%5BA%5D_%7B0%7D-kt)
Therefore a plot of ln[A] vs t should give a straight line with a slope = -k
Hence, ln[reactant] vs time should be plotted for a first order reaction.
The answer is b since on a it says H2 but on the right side there is no H idk if you forgot to put H there so im guessing b. Have a good day