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?

1 Answer
Mar 31, 2015

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"#