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

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

Final answer:

The enthalpy change (ΔH) for the reaction 3A+6B→3C+9D is three times the original ΔH, given that the original reaction A+2B→C+3D has a ΔH of 161 kJ. Thus, ΔH for the related reaction is 483 kJ.

Explanation:

The reaction given: A+2B→C+3D, has an enthalpy change (ΔH) of 161 kJ per stoichiometric coefficient of the reaction. Therefore, when you multiply the reaction by three to give 3A+6B→3C+9D, the ΔH value will also be multiplied by three. Thus, the ΔH for the reaction 3A+6B→3C+9D is 161 kJ x 3, which is equal to 483 kJ.

This is because ΔH for a reaction is directly proportional to the number of moles or the quantity of reactants and products involved as shown by Hess's law. For instance, if you have twice or three times the quantity of reactants, you would release or absorb twice or three times that amount of energy, respectively.

Moreover, the enthalpy change of a reaction can also be determined by the enthalpies of formation of the reactants and products or the bond energies depending on the details available. But in this case, the information given directly allows for a simple multiplication to find the answer.

Learn more about Enthalpy Change here:

brainly.com/question/32882904

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