answer: the value of enthalpy change will be -152.80 kj.
solution:
pressure = 1atm , temperature =
= 298 k ![)0^oc=273kelvins](/tex.php?f=)0^oc=273kelvins)
![o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol](/tex.php?f=o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol)
![\delta h_{no_2}=33.85kj/mol](/tex.php?f=\delta h_{no_2}=33.85kj/mol)
![\delta h_{o_3}=142.2kj/mol](/tex.php?f=\delta h_{o_3}=142.2kj/mol)
![\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})](/tex.php?f=\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3}))
![\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol](/tex.php?f=\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol)
given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.
![v_{no}=12.00 l,v_{o_3}=8.50 l](/tex.php?f=v_{no}=12.00 l,v_{o_3}=8.50 l)
![pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k](/tex.php?f=pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k)
![n_{no}=0.49 moles](/tex.php?f=n_{no}=0.49 moles)
![pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k](/tex.php?f=pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k)
![n_{o_3}=0.34 moles](/tex.php?f=n_{o_3}=0.34 moles)
according to reaction 3 moles no are reacting with 1 mol of
,then 0.49 mol of no will react with
moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and
is an excessive reagent.
enthalpy change for the reaction between the given volumes of the gases is :
![\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol](/tex.php?f=\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol)