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Sergeeva-Olga [200]
3 years ago
11

The enthalpies of formation of the compounds in the combustion of methane,

Chemistry
2 answers:
Naya [18.7K]3 years ago
5 0

Answer:

The amount of heat released by the combustion of 2 moles of methane = 1605.1 kJ

Option 3 is correct.

ΔH°(combustion of 2 moles of methane)

= -1,605.1 kJ

Explanation:

The balanced equation for the combustion of methane is presented as

CH₄ + 2O₂ → CO₂ + 2H₂O

The heat of formation for the reactants and products are

CH₄ (g): ΔHf = –74.6 kJ/mol;

CO₂ (g): ΔHf = –393.5 kJ/mol;

and H₂O(g): ΔHf = –241.82 kJ/mol.

ΔHf for O₂ = 0 kJ/mol

ΔH°rxn for the combustion of methane is given as

ΔH°rxn = ΣnH°(products) - ΣnH°(reactants)

ΣnH°(products) = (1×-393.5) + (2×-241.82)

= -877.14 kJ/mol

ΣnH°(reactants) = (1×-74.6) + (2×0)

= -74.6 kJ/mol

ΔH°rxn = -877.14 - (-74.6) = -802.54 kJ/mol

For 2 moles of methane, the heat of combustion = 2 moles × -802.54 kJ/mol

= -1,605.08 kJ = -1,605.1 kJ

Hope this Helps!!!

inn [45]3 years ago
4 0

Answer:

C, -1,605.1 kJ

Explanation:

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Explanation : Given,

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The balanced chemical reaction is:

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Initial moles     0.1116       0.0461     0.08978

At eqm.       (0.1116-0.0461)    0       (0.08978+0.0461)

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Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

pH=9.31+\log (\frac{0.1359}{0.0655})

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Answer : The balanced chemical equation in a acidic solution is,

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Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

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Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion (H^+) at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

BrO_3^-(aq)+Sb^{3+}(aq)\rightarrow Br^-(aq)+Sb^{5+}(aq)

The oxidation-reduction half reaction will be :

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-

  • First balance the main element in the reaction.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-

  • Now balance oxygen atom on both side.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-\rightarrow Br^-+3H_2O

  • Now balance hydrogen atom on both side.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}

Reduction : BrO_3^-+6H^+\rightarrow Br^-+3H_2O

  • Now balance the charge.

Oxidation : Sb^{3+}\rightarrow Sb^{5+}+2e^-

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The charges are not balanced. Now multiplying oxidation reaction by 3 and then adding both equation, we get the balanced redox reaction.

Oxidation : 3Sb^{3+}\rightarrow 3Sb^{5+}+6e^-

Reduction : BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O

The balanced chemical equation in acidic medium will be,

BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)

The sum of the coefficients = 1 + 6 + 3 + 1 + 3 + 3

The sum of the coefficients = 17

7 0
3 years ago
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