Question

Hess's Law Calculate the Value of H for: 3NO2(g)+2H2O(l) = 2HNO3(aq)+NO(g)

Using this information: 2NO(g)+O2(g) = 2NO2(g) , H=-116 kJ

2N2(g)+5 O2(g)+2H2O(l) = 4HNO3(aq) , H= -256kJ

N2(g) + O2(g) = 2NO(g) , H=+183kJ

Answer 1

Ans. Reaction 1:       3 NO₂(g) + 2 H₂O(l) = 2 HNO₃(aq) + NO(g)

Reaction 2: 2 NO(g) + O₂(g) --------> 2 NO₂(g)                         ; dH = -116 kJ

Reaction 3: 2 N₂(g) + 5 O₂(g) + 2 H₂O(l) ----> 4 HNO₃(aq)       ; dH = - 256 kJ

Reaction 4: N₂(g) + O₂(g) --------> 2 NO(g)                                 ; dH = +183 kJ

Note the following points-

I. When a reaction is reversed, the sign of dH also reversed.

II. If the reaction is multiplied by a factor, the dH value is multiplied by same factor, too.

# Reaction 1 can be written as the sum of –

            (1.5 x Reverse of Rxn 2) + (0.5 x Rxn 3) + (Reverse of Rxn 4) as follow-

            3 NO₂(g) ------------------------> 3 NO(g) + 1.5 O₂(g)               ; dH = + 174 kJ

(+)      N₂(g) + 2.5 O₂(g) + H₂O(l) -----> 2 HNO₃(aq)                             ; dH = - 128 kJ

(+)      2 NO(g) ------------------------> N₂(g) + O₂(g)                           ; dH = - 183 kJ

3 NO₂(g) + 2 H₂O(l) = 2 HNO₃(aq) + NO(g)                            ; dH = ?

Using Hess’s Law-

dH of reaction 1 = 174 kJ + (-128 kJ) + (-183 kJ)

                        = - 137 kJ