I think it is B, but i am not sure
An acid-base reaction or a neutralization reaction is a <u>chemical reaction that occurs between an acid and a base producing a salt and water</u>. The acids and bases can be strong or weak depending on their degree of ionization in water.
Butyric acid is a weak acid and in water it is ionized in the following way, loosing a proton (H+):
HC4H7O2 (aq) ⇆ H+ (aq) + C4H7O2- (aq)
On the other hand, potassium hydroxide is a strong base, so it will be completely ionized in water:
KOH(aq) → K+(aq) + OH-(aq)
Then the <u>net acid-base reaction</u> between butyric acid and KOH is:
HC4H7O2 (aq) + OH- (aq) ⇆ H2O + C4H7O2- (aq)
It is valid to consider only the OH- produced from the ionization of KOH in water since, as mentioned, this molecule is completely ionized. Also, we do not include the K + in the net equation since it is a spectator ion, it does not undergo chemical changes.
<span>
</span><span> average reaction rate </span><span>= change in concentration / change in time
by putting values we have
= (1.00M - 0.987M) / (4.00s - 0.00s)
= 3.25x10^-3 mol/Lsec
this is our conclusion
hope this helps</span>
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