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

What would the enthalpy change for the following reaction be?

4 NO2(g) ------- 2 N2(g) + 4 O2(g) ΔH = ???

a. +66.4 kJ
b. -66.4 kJ
c. +132.8 kJ
d. -132.8 kJ
e. -33.4 kJ

Respuesta :

joseaaronlara

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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