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krek1111 [17]
3 years ago
12

Which equation represents a conservation of atoms? Group of answer choices LaTeX: 4Fe\:+3O_2\:\longrightarrow\:2Fe_2O_3 4 F e +

3 O 2 ⟶ 2 F e 2 O 3 LaTeX: 2Fe\:+2O_2\:\longrightarrow\:Fe_2O_3 2 F e + 2 O 2 ⟶ F e 2 O 3 LaTeX: 4Fe\:+2O_2\:\longrightarrow\:2Fe_2O_3 4 F e + 2 O 2 ⟶ 2 F e 2 O 3 LaTeX: 2Fe\:+\:3O_2\:\longrightarrow\:Fe_2O_3
Chemistry
1 answer:
slamgirl [31]3 years ago
3 0

Answer : The equation represents a conservation of atoms is:

4Fe+3O_2\rightarrow 2Fe_2O_3

Explanation :

Law of conservation of matter : It states that matter can neither be created nor be destroyed but it can only be transformed from one form to another form.

The balanced chemical reaction always follow the law of conservation of matter.

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

(1) The given chemical reaction is,

4Fe+3O_2\rightarrow 2Fe_2O_3

This reaction is a balanced chemical reaction in which number of atoms of iron and oxygen are equal on the both side of the reaction. So, this reaction obey the law of conservation of matter.

(2) The given chemical reaction is,

2Fe+2O_2\rightarrow 2Fe_2O_3

This reaction is an unbalanced chemical reaction in which number of atoms of iron and oxygen are not equal on the both side of the reaction. So, this reaction does not obey the law of conservation of matter.

(3) The given chemical reaction is,

4Fe+2O_2\rightarrow 2Fe_2O_3

This reaction is an unbalanced chemical reaction in which number of atoms of oxygen are not equal on the both side of the reaction. So, this reaction does not obey the law of conservation of matter.

Hence, the equation represents a conservation of atoms is:

4Fe+3O_2\rightarrow 2Fe_2O_3

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Answer:

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Explanation:

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Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC
Salsk061 [2.6K]

<u>Answer:</u> The \Delta G for the reaction is 54.6 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

  • To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

  • The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm

Putting values in above equation, we get:

K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92

  • To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 22.92

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol

Hence, the \Delta G for the reaction is 54.6 kJ/mol

7 0
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