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maria [59]
3 years ago
7

Which one of the following is an oxidation-reduction reaction?

Chemistry
2 answers:
kupik [55]3 years ago
7 0

Answer:

CH4 + 2 O2 --> CO2 + 2 H2O

Explanation:

CH4 + 2 O2 --> CO2 + 2 H2O is the only reaction where an element (oxygen) undergoes a change in oxidation state. In this reaction oxygen changes disproportionately to O⁻². That is ...

O₂ → CO₂ + 4e⁻ ==> oxidation

<u>O₂ + 4e⁻  →  H₂O ==> reduction </u>

2O₂ + 4e⁻  →  CO₂ + H₂O + 4e⁻  ==> Net oxidation-reduction

=>  4e⁻ gained by one mole O₂ in formation of CO₂ = 4e⁻ lost by the other mole O₂ in forming H₂O.

Then...

Including CH₄ (whose elements do not undergo changes in oxidation states) requires doubling reaction to balance by mass thus giving ...

2CH₄ + 2O₂ + 8e⁻  →  2CO₂ + 2H₂O + 8e⁻

Cancelling 8 reduction electrons on left with 8 oxidation electrons on right gives...

2CH₄ + 2O₂  →  2CO₂ + 2H₂O

ivann1987 [24]3 years ago
4 0

Answer:

CH₄ + 2O₂ ⟶ CO₂ + 2H₂O  

Explanation:

To identify an oxidation-reduction reaction, you must determine the oxidation number of every atom involved in the reaction and see if it changes.

The only reaction where two elements change oxidation number is the oxidation of methane.

Here's the oxidation number of every atom involved:.

\stackrel{\hbox{-4}}{\hbox{C}}\stackrel{\hbox{+1}}{\hbox{H}}_{4} +\stackrel{\hbox{0}}{\hbox{O}}_{2} \, \longrightarrow \, \stackrel{\hbox{+4}}{\hbox{C}}\stackrel{\hbox{-2}}{\hbox{O}}_{2} + \stackrel{\hbox{+1}}{\hbox{H}}_{2}\stackrel{\hbox{-2}}{\hbox{O}}

We see that some elements change oxidation numbers.

C:  -4 ⟶ +4; increase in oxidation number = oxidation

O:  0 ⟶  -2; decrease in oxidation number = reduction

H: +1 ⟶   +1; no change.

The reaction is an oxidation-reduction reaction, because carbon is oxidized, and oxygen is reduced.

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Given that:

  • The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
  • normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K  = 272.6 K
  • volume  = 250 mL = 0.250 L
  • Mass of butane = 0.8 g
  • the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}

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P1 and P2 correspond to the temperature at T1 and T2.

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replacing the values into the given equation, we have;

\mathbf{In \dfrac{P_2}{1\  atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}

\mathbf{In \dfrac{P_2}{1\  atm} =-(0.852053785)}

\mathbf{P_2=0.427 \ atm}

As such, at -22° C; the vapor pressure = 0.427 atm

Now, using the ideal gas equation:

PV = nRT

where:

  • P = Pressure
  • V = volume
  • n = number of moles of butane
  • R = universal gas constant
  • T = temperature

∴

Making (n) the subject of the formula:

\mathbf{n = \dfrac{PV}{RT}}

\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}

\mathbf{n =0.00518 mol}

We all know that the standard molecular weight of butane = 58.12 g/mol

∴

Using the relation for the number of moles which is:

\mathbf{number \  of \  moles = \dfrac{mass}{molar mass}}

mass = 0.00518 mole × 58.12 g/mol

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∴

The mass of butane in the flask = 0.301 g

But the mass of the butane present as a liquid in the flask is

= 0.8 g - 0.301 g

= 0.499 g

In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g

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