Question

NItrogen in air reacts at high temperature to form NO2 according to the reaction:

N2 + 2 O2 ? 2 NO2

Draw structures for the reactants and products indicating the number of single, double and triple bonds.

Determine the ?H for the reaction using table 7.1.

[Hint: one NO2 has one [N-O] bond and one [N=O] bond]



A.

+115 kcal

B.

-78 kcal

C.

+78 kcal

D.

+7 kcal

E.

none of these

Answer 1

Answer:

The correct answer is option E.

Explanation:

Structures for the reactants and products are given in an aimage ;

Number of double bonds in oxygen gas molecule = 1

Number of double bonds in nitro dioxide gas molecule = 1

Number of single bond in in nitro dioxide gas molecule = 1

Number of triple bonds in nitrogen gas molecule = 1

$N_2+2O_2\rightarrow 2NO_2,\Delta H=?$

$\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]$

$\Delta H_{f,NO_2}=33.18 kJ/mol$

$\Delta H_{f,N_2}=0$ (pure element)

$\Delta H_{f,O_2}=0$ (pure element )

$\Delta H=2 mol\times 33.18 kJ/mol=66.36kJ=15.86 kcal$

The enthalpy of the given reaction is 15.86 kcal.