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ikadub [295]
3 years ago
14

Using the following thermochemical data, what is the change in enthalpy for the following reaction: 3H2(g) + 2C(s) + ½O2(g) → C2

H5OH(l) C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l), ΔH = –1367 kJ/mol C(s)+O2(g)→CO2(g), ΔH = –393.5 kJ H2(g)+½O2(g)→H2O(l), ΔH = –285.8 kJ
Chemistry
2 answers:
sweet [91]3 years ago
7 0
It would be -277.6 KJ/mol
Sergio [31]3 years ago
5 0

Answer:

The change in enthalpy for the formation of ethanol is -276.5 kJ/mol

Explanation:

The given reactions are:

C2H5OH(l)+3O2(g)\rightarrow 2CO2(g)+3H2O(l) ----\Delta H_{1}=-1367kJ/mol

C(s)+O2(g)\rightarrow CO2(g)-----\Delta H_{2}=-393.5 kJ/mol

H2(g)+ 1/2O2(g)\rightarrow H2O(l)-----\Delta H_{3}=-285.8 kJ/mol  

The required reaction involves the formation of C2H5OH from C, H2 and O2:

3H2(g) + 2C(s) + 1/2O2(g) \rightarrow  C2H5OH(l) -----\Delta H_{rxn}=?

This can be obtained by reversing reaction (1), multiplying equation (2) by 2, multiplying equation (3) by 3 and then adding the corresponding ΔH values

\Delta H_{rxn}= -\Delta H_{1}+ 2(\Delta H_{2})+3(\Delta H_{3})

\Delta H_{rxn}= 1367 + 2(-393.5)+3(-285.5)=-276.5 kJ/mol

   

   

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Answer:

(a)²⁴²₉₄Pu ⇒ ⁴₂He + ²³⁸₉₂U ⇒ ²³⁸₉₂U + ⁰₀γ

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Explanation:

In the nuclear reaction, it was stated that plutonium-242 decayed firstly to uranium-238 and alpha, and lastly to a stable uranium-238 by emitting a gamma ray. The balanced equation for the nuclear reactions is shown below:

(a)²⁴²₉₄Pu ⇒ ⁴₂He + ²³⁸₉₂U ⇒ ²³⁸₉₂U + ⁰₀γ

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Therefore:

E = (4.136*10^-15)*(299792458)/0.02757*10^-9 = 44974.31 eV

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tino4ka555 [31]

Answer:

ΔH°rxn = 54.08 kJ

Explanation:

Let's consider the following equations.

a) ClO(g) + O₃(g) ⇄ Cl(g) + 2 O₂(g)                     ΔH°rxn = –29.90 kJ

b) 2 O₃(g) ⇄ 3 O₂(g)                                            ΔH°rxn = 24.18 kJ

We have to determine the value of heat of reaction for the following reaction: Cl(g) + O₃(g) ⇄ ClO(g) + O₂(g)

According to Hess's law, the enthalpy change in a chemical reaction is the same whether the reaction takes place in one or in several steps. That means that we can find the enthalpy of a reaction by adding the corresponding steps and adding their enthalpies. According to Lavoisier-Laplace's law, if we reverse a reaction, we also have to reverse the sign of its enthalpy.

Let's reverse equation a) and add it to equation b).

-a) Cl(g) + 2 O₂(g) ⇄ ClO(g) + O₃(g)                    ΔH°rxn = 29.90 kJ

b) 2 O₃(g) ⇄ 3 O₂(g)                                            ΔH°rxn = 24.18 kJ

-------------------------------------------------------------------------------------------------

Cl(g) + 2 O₂(g) + 2 O₃(g) ⇄ ClO(g) + O₃(g) + 3 O₂(g)

Cl(g) + O₃(g) ⇄ ClO(g) +O₂(g)

ΔH°rxn = 29.90 kJ + 24.18 kJ = 54.08 kJ

3 0
3 years ago
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