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denis23 [38]
3 years ago
13

What is the overall enthalpy of reaction for the equation shown below?

Chemistry
1 answer:
Rudiy273 years ago
4 0

Answer:

ΔH₁₂ = -867.2 Kj

Explanation:

Find enthalpy for 3H₂ + O₃ => 3H₂O given ...

2H₂ + O₂ => 2H₂O      ΔH₁ = -483.6 Kj

        3O₂ => 2O₃        ΔH₂ = + 284.6 Kj

_____________________________

3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O       (multiply by 3 to cancel O₂)

6H₂ + 3O₂ => 6H₂O        ΔH₁ = 3(-483.6 Kj) = -1450.6Kj

          2O₃ => 3O₂           ΔH₂ = -284.6Kj              (reverse rxn to cancel O₂)

_______________________________

6H₂ + 2O₃ => 6H₂O         ΔH₁₂ = -1735.2 Kj       (Net Reaction - not reduced)

________________________________

divide by 2 => target equation (Net Reaction - reduced)

3H₂ + O₃ => 3H₂O            ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj    

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Answer:

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Explanation:

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here is your answer >>>

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Solution for A gas has a volume of 340.0 mL at 45.90 degree celsius. What is the new temperature of the gas, in kelvin, if the volume increased to 550.0 mL.

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Hello!

i. In this case, since the given chemical reaction is exothermic due to the negative change in the enthalpy of reaction, we infer that according to the mentioned principle, by lowering the temperature the reaction will shift rightwards and therefore the yield is increased; thus, you need a lower temperature than the specified.

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