Question

Considered the reaction: N2(g) + 2NO2(g) for which (delta H) = +67.6KJ. What is the enthalpy change in KJ for the production of 23.0 g NO2?

Answer 1

The enthalpy change is +16.9kJ.

I'll assume you mean:

`N_(2(g))+2O_(2(g))rarr2NO_(2(g))`

`DeltaH=+67.6"kJ"`

The formation of 2 moles of `NO_2` results in `67.6"kJ"` being absorbed from the surroundings.

Now convert moles to grams, I will use approximate `A_r` values. You should use the ones you are given.

The `M_r` of `NO_2=[14.0+(2xx16.0)]=46.0`

So 1 mole weighs 46.0g

So `2xx46.0=92.0"g"` `rarr67.6"kJ"`

So `1"g"rarr(67.6)/(92.0)"kJ"`

So `23.0"g"rarr(67.6)/(92.0)xx23.0=16.9"kJ"`