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34kurt
3 years ago
12

Consider the exothermic reaction

Chemistry
1 answer:
alisha [4.7K]3 years ago
4 0

Answer:

The answer to your question is -2855 J

Explanation:

Reaction

                     2C₂H₆  +  7O₂   ⇒   4CO₂  +  6H₂O

Formula

Heat of reaction = ΔHrxn = ΣΔHrxn products - ΣΔHrxn reactants

Substitution

ΔHrxn = { 4(-393.5) + 6(-241.8)} - {2(-84.7) + 7(0)}

ΔHrxn = {-1574 -1450.8} - {-169.4}

ΔHrxn = -3024.8 + 169.4

ΔHrxn = -2855.4 J

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Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
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See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

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now

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so

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now

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reducing agents are Cr and N2

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lower the reduction potential , stronger the reducing agent

so

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reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

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now

oxidizing agents :

Au+ > Cr+3

reducing agents :

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3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0
3 years ago
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