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Arada [10]
2 years ago
14

2N2 + 3H2 → 2NH3

Chemistry
1 answer:
denis23 [38]2 years ago
3 0

There are two N≡N bonds and three H–H  bonds are in reactants.

Given:

The reaction between nitrogen gas and hydrogen gas.

2N_2+3H_2\rightarrow 2NH_3

To find:

Bonds on the reactant side

Solution:

2N_2+3H_2\rightarrow 2NH_3

Reactants in the reaction = N_2, H_2

The bond between nitrogen atoms in single N_2 molecule  = N≡N (triple bond)

Then in two N_2  molecules = 2 N≡N (triple bonds)

The bond between hydrogen atoms in single H_2 molecule = H-H (single bond)

Then in three H_2  molecules = 3  H-H (single bonds)

Product in the reaction =NH_3

The bonds between nitrogen and hydrogen atoms in single NH_3 molecule = 3 N-H (single bond)

Then in two NH_3  molecules = 6  N-H (single bonds)

So, there are two N≡N bonds and three H–H bonds are in reactants.

Learn more about reactants and products here:

brainly.com/question/21517037?referrer=searchResults

brainly.com/question/20602904?referrer=searchResults

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For the following electrochemical reaction: Al3+(aq) + 3e -&gt; Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -&gt; 2F (aq) Calculate
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<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V

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

Aluminium will undergo oxidation reaction and will get oxidized.

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

To calculate the E^o_{cell} of the reaction, we use the equation:

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