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Elena L [17]
3 years ago
12

Use the information provided to determine ΔH°rxn for the following reaction: ΔH°f (kJ/mol) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2

(g) ΔH°rxn = ? Fe2O3(s) -824 Fe3O4(s) -1118 CO(g) -111 CO2(g) -394
Chemistry
2 answers:
Crazy boy [7]3 years ago
4 0

Answer:

ΔH°rxn = - 47 KJ/mol

Explanation:

  • 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g)

∴ ΔH°f  Fe2O3(s) = - 824 KJ/mol

∴ ΔH°f Fe3O4(s) = - 1118 KJ/mol

∴ ΔH°f CO(g) = - 111 KJ/mol

∴ ΔH°f CO2(g) = - 394 KJ/mol

  • ΔH°rxn = ∑νiΔH°fi

⇒ ΔH°rxn = [ΔH°f CO2(g) + 2 ΔH°f Fe3O4(s)] - [ΔH°f CO(g) + 3 ΔH°f Fe2O3(s)]

⇒ ΔH°rxn = [- 394 KJ/mol + 2(- 1118 KJ/mol)] - [(- 111 KJ/mol) + 3(- 824 KJ/mol)

⇒ ΔH°rxn = - 2630 KJ/mol - (- 2583 KJ/mol)

⇒ ΔH°rxn = - 47 KJ/mol

zepelin [54]3 years ago
3 0

Answer:

ΔH°rxn = -47 kJ

Explanation:

Using Hess´s law for the reaction:

3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ,

the ΔH°rxn will be given by the expression:

ΔH°rxn kJ  =   2ΔHºf(Fe3O4)  + ΔHºf(CO2)   - ( 3ΔHºf(Fe2O3) + ΔHºf(CO) )

=  2(-1118) + (-394) - ( 3( -824 ) + ( -111 ) )

= - 47 kJ

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spayn [35]

Explanation:

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{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

6 0
3 years ago
1. When this equation is balanced, Fe + 02 -.--> Fe203, what is the
Vilka [71]

Fe + O2 → Fe2O3

After balancing the eq.

4Fe + 3O2 → 2Fe2O3

Hope this will help u mate :)

4 0
3 years ago
Which of the following compounds is a gas at room temperature? A : CH3CH2OH B : CH3CH2CH2CH2CH2CH2CH3 C : CH3CH2CH3 D : HOCH2CH2
attashe74 [19]

Answer:

C. CH_3CH_2CH_3

Explanation:

Molecules with the stronger intermolecular forces are pulled tightly together  to form solid at higher temperatures and that's why the freezing point is higher.

Also, molecules with the stronger intermolecular force have greater interaction between the molecules and thus on heating do not boil easily and have high boiling point also.

Thus, melting point and boiling point increases with increase in number of carbon atoms and also increase in intermolecular forces (like hydrogen bonding, if present).

Thus, the compound which is gas at room temperature is CH_3CH_2CH_3 <u>because it has least number of carbon atoms and absence of hydrogen bonding.</u>

<u></u>

8 0
3 years ago
49.9 ml of a 0.00292 m stock solution of a certain dye is ddiluted to 1.00 L. the diluted solution has an absorbance of 0.600. w
inna [77]

Complete Question

49.9 ml of a 0.00292 m stock solution of a certain dye is diluted to 1.00 L. the diluted solution has an absorbance of 0.600. what is the molar absorptivity coefficient of the dye

Answer:

The  value is  \epsilon  =  4118.1 \  M^{-1} cm^{-1}  

Explanation:

From the question we are told that

   The volume of the stock solution is  V_1   =  49.9 mL  =  0.0499 \  L  

   The concentration of the stock solution is  C_1  =  0.00292 \  M

   The volume of the diluted solution is  V_2 =  1.00 \  L

   The absorbance is  A =  0.600

Generally the from the titration equation we have that

         C_1 * V_1 =  C_2 * V_2

=>      0.00292  * 0.0499 =  C_2 * 1

=>     C_2 = 0.0001457 \  M

Generally from  Beer's law we have that

      A  =  \epsilon  * l  *  C_2

=>   \epsilon  =  \frac{A}{ l  *  C_2 }

Here  l is the length who value is  1 cm because the unit of  molar  absorptivity coefficient of the dye is M^{-1} *  cm^{-1}

So

            \epsilon  =  \frac{0.600}{ 1   * 0.0001457   }  

=>       \epsilon  =  4118.1 \  M^{-1} cm^{-1}  

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