Which answer do you this is correct?

Which answer do you this is correct?


100%sorry if i’m wronghope this good luck: )

answer: the value of enthalpy change will be -152.80 kj.


pressure = 1atm , temperature = 25^oc= 298 k )0^oc=273kelvins

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

\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})

\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

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

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

according to reaction 3 moles no are reacting with 1 mol of o_3 ,then 0.49 mol of no will react with \frac{1}{3}\times 0.49 moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and o_3 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

The fourth one: Gain of an electron

The fourth one: Gain of an electron

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