[填空题]
The standard enthalpy of fovd7q 6nr2(f bcrmation of gaseous ni //iu et5wa9e3plq*;rsk3 fo btric oxide,ΔH$_f^θ$(NO(g)),is +90.3kJmol$^{-1}$.
1.Write a reaction equation, including state symbols, corresponding to the standard enthalpy of formation of gaseous nitric oxide.
2.Calculate the bond enthalpy of nitric oxide, using the value of ΔH$_f^θ$(NO(g)) above and section 11 (Section 12- 2025 Syllabus) of the data booklet.
E(N≡O)= (kJ mol$^{-1}$)
3.In the Ostwald process, nitric oxide (NO) is reacted with oxygen and then water to produce aqueous nitric acid($HCO_3$).The overall equation and enthalpy change of this reaction are given below.
4NO(g)+3$O_2$(g)+2$H_2O$(l)→4$HNO_3$(aq)
ΔH$_1$=-576.4kJ mol$^{-1}$
Calculate the enthalpy of formation of aqueous nitric acid, ΔH$_f^θ$($HNO_3$(aq)),using the values of ΔH$_f^θ$(NO(g))=+90.3kJ mol$^{-1}$,and ΔH$_1$ and section 12 (Section 13- 2025 Syllabus) of the data booklet.
ΔH$_f^θ$($HNO_3$(aq))= (kJ mol$^{-1}$)
4.The nitric oxide used in the Ostwald process is formed by reacting gaseous ammonia ($NH_3$) with oxygen. The equation and enthalpy change of this reaction are given below.
4$NH_3$(g)+5$O_2$(g)→4$NO$(g)+6$H_2O$(g)
ΔH$_2$=-906.0kJ mol$^{-1}$
The yield of the Ostwald process may be lower than expected if some of the nitric oxide product reacts with the remaining ammonia. The equation for this side reaction is shown.
4$NH_3$(g)+6$NO$(g)→5$N_2$(g)+6$H_2O$(g)
4.1Calculate the enthalpy change for the side reaction, Δ$H_{side}$,using the values of ΔH$_f^θ$(NO(g)) and ΔH$_2$.
Δ$H_{side}$=- (kJ mol$^{-1}$)
2.Suggest how the value of ΔH$_2$ would differ if water were produced in the liquid phase.
The value of ΔH$_2$ would be negative
