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olchik [2.2K]
3 years ago
8

Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

(l) ΔH°f = -1411 kJ C(s) + O2(g)--> CO2(g) ΔH°f = -393.5 kJ H2(g) + 1/2 O2(g)--> H2O(l) ΔH°f = -285.8 kJ
Chemistry
1 answer:
Brums [2.3K]3 years ago
7 0

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 C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.5kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ

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

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.5kJ)=-787kJ

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.8kJ)=-571.6kJ

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)

\Delta H=52.4kJ

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

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Which of the following acids is the WEAKEST? The acid is followed by its Ka value.HF, 3.5 × 10^-4HNO2, 4.6 × 10^-4HCN, 4.9 × 10^
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Answer:

HCN

Explanation:

There are several factors that can tell us when an acid is stronger than another, which are the following:

1. The Polarity of the X - H Bond

2. Size of the X atom.

3. Charge on the acid.

4. Oxidation state of the central atom

5. Values of Ka and pKa.

From all of this factors, we can see that the exercise is already giving us values of Ka, and also we have different types of acid, not only with the form H - X

So, we will base the force of an acid by it's pKa value.

The pKa value which is calculated with the expression:

pKa = -logKa

pKa is a value that indicates how strong is the acid you are working with. This value can vary depending on factors such the charge of the acid. I f this charge can be easily distributed in resonance structures, the compound is more acidic, and therefore the pKa value is lower.

The lower the pKa, the more acidic the compound is.

So calculating the pKa on every structure we have:

HF: pKa = -log(3.5x10^-4) = 3.46

HNO2: pKa = -log(4.6x10^-4) = 3.34

HCN: pKa = -log(4.9x10^-10) = 9.31

HCOOH: pKa = -log(1.8x10^-4) = 3.74

HClO2: -log(1.1x10^-2) = 1.96

So according to all this values, we can conclude that the weakest acid is the HCN

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