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3 years ago

N2(g) + 2 O2(g) ------- 2 NO2(g) ΔH = +66.4 kJ

Chemistry

1 answer:

SpyIntel [72]3 years ago

5 0

Answer:

The answer to your question is the letter d. -132.8 kJ

Explanation:

To solve this problem, just observe that the ΔH is originally equal to +66.4 kJ.

If we change the order of the reaction, ΔH changes its sign but the value is the same.

Finally, if we duplicate the number of moles, also the ΔH increases twice.

For this problem

ΔH = 2(-66.4 kJ)

ΔH = - 132.8 kJ

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Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energ

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Explanation:

We are given:

$E^o_{(NO_2^-/NH_4)}=-0.41V\\E^o_{(O_2/H_2O)}=0.82V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, oxygen will undergo reduction reaction will get reduced.

$NH_4$ will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

The half reactions follows:

Oxidation half reaction: $NH_4\rightarrow NO_2^-+6e^-$ ( × 4)

Reduction half reaction: $O_2+4e^-\rightarrow 2H_2O$ ( × 6)

Overall reaction: $4NH_4+6O_2\rightarrow 4NO_2^-+12H_2O$

We are given:

Molecules of $NH_4$ = 28

By Stoichiometry of the reaction:

4 molecules of $NH_4$ reacts with 6 molecules of oxygen gas

So, 28 molecules of $NH_4$ will react with = $\frac{6}{4}\times 28=42$ molecules of oxygen gas

Hence, the molecules of oxygen gas that will be reduced to water are 42 molecules

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