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

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

43.452

step-by-step explanation:

284 times .153 = 43.452