Question

27. Use Hess’ Law to calculate the enthalpy change for the decomposition of dinitrogen pentaoxide?

Answer 1

`DeltaH=-28.4kJ`

Explanation:

We must construct the overall decomposition reaction from the more elementary reactions.

We need `2N_2O_5` on the left hand side so we need to reverse and double the second reaction. We write this as.

`2N_2O_5+2H_2Orarr4HNO_3`, `DeltaH=-2*76.6 = -153.2kJ`

We need `2N_2` on the right-hand side so we need to reverse and quadruple the third reaction.

`4HNO_3rarr2N_2+6O_2+2H_2`, `DeltaH=-4*(-174.1)=696.4kJ`

Now we need to consume the two moles of diatomic hydrogen generated from this last reaction. We can double the first reaction for this.

`2H_2+O_2rarr2H_2O`, `DeltaH=2*(-285.8)=-571.6kJ`

Now note that all of these reactions add up to the target reaction

`2N_2O_5+2H_2Orarr4HNO_3`
`4HNO_3rarr2N_2+6O_2+2H_2`
`2H_2+O_2rarr2H_2O`

sum to

`2N_2O_5rarr2N_2+5O_2`

Because enthalpy is a state variable, Hess's Law says the enthalpy change for this reaction is

`DeltaH=-153.2+696.4-571.6=-28.4kJ`.