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Consider the following generic reaction: A+2B→C+3D, with ΔH = 161 kJ . Determine the value of ΔH for each of the following related reactions.3A+6B→3C+9D

Respuesta :

Answer:

ΔH=483 kJ

Explanation:

Its given that

For the reaction A+2B→C+3D has  ΔH = 161 kJ.

As the ΔH is positive this is an endothermic reaction where we supply energy.

It is given that we supposed to find ΔH for the following equation.

3A+6B→3C+9D    ΔH=?

So we observe that

3*(A+2B)→3*(C+3D)

So the whole reaction is rised with a coefficient of 3 hence the ΔH would be three times the original one.

ΔH=3*161 kJ

ΔH=483 kJ

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