Question

The following are some enthalpy values for reactions between nitrogen and oxygen: N2(g) + O2(g) → 2NO(g) ∆H = 180.4 kJ mol−1 N2(g) + 2O2(g) → 2NO2(g) ∆H = 66.4 kJ mol−1 Use these values to calculate ∆H for the reaction: NO(g) + ½O2(g) → NO2(g)?

Answer 1

I make it....`-114.0*kJ*mol^-1`

Explanation:

And so we gots....

`"(i) " N_2(g) + O_2(g) rarr 2NO(g);DeltaH_"rxn i"^@=+180.4*kJ*mol^-1`

`"(ii) " N_2(g) + 2O_2(g) rarr 2NO_2(g);DeltaH_"rxn ii"^@=+66.4*kJ*mol^-1`

We want...

`"(iii) "NO(g) + 1/2O_2(g) rarr NO_2(g);`

But `(iii)=1/2xx{(ii)-(i)}` ......

`"(iii) "1/2N_2(g) + O_2(g) +NO(g)rarr NO_2(g)+1/2N_2(g) + 1/2O_2(g)`....

`"(iii) "NO(g)+ 1/2O_2(g)rarr NO_2(g)`

...which is PRECISELY the expression whose enthalpy change we want to evaluate...

And so...`DeltaH_"rxn (iii)"=1/2DeltaH_"rxn (ii)"^@-DeltaH_"rxn (i)"^@`

`={+66.4-180.4}*kJ*mol^-1=??*kJ*mol^-1`