Question

The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes,

(1) NO(g) + NO
2
(g)
→
N
2
O
3
(g) ;
Δ
H
o
r
x
n
= -39.kJ

(2) NO(g) + NO
2
(g) + O
2
(g)
→
N
2
O
5
(g) ;
Δ
H
o
r
x
n
= -112.5 kJ

(3) 2NO
2
(g)
→
N
2
O
4
(g) ;
Δ
H
o
r
x
n
= -57.2 kJ

(4) 2NO(g) + O
2
(g)
→
2NO
2
(g) ;
Δ
H
o
r
x
n
= -114.2 kJ

(5) N
2
O
5
(s)
→
N
2
O
5
(g) ;
Δ
H
o
s
u
b
l
= 54.1 kJ

Calculate the heat of reaction for N
2
O
3
(g) + N
2
O
5
(s)
→
2N
2
O
4
(g)

Δ
H = _ _ _ _ _ kJ

Answer 1

The heat of the given reaction is -23.0 kJ.

Step-by-step explanation:

We are given with ;

`NO(g) + NO_2(g)\rightarrow N_2O_3(g) \Delta H_(1,rxn)=-39.kJ`..[1]

`NO(g) + NO_2(g) + O_2(g)\rightarrow N_2O_5(g), \Delta H_(2,rxn) = -112.5 kJ`..[2]

`2NO_2(g) \rightarrow N_2O_4(g) ,\Delta H_(3,rxn) = -57.2 kJ`..[3]

`2NO(g) + O_2(g)\rightarrow 2NO_2(g), \Delta H_(4,rxn) = -114.2 kJ`..[4]

`N_2O_5(s)\rightarrow N_2O_5(g) ,\Delta H_(5,sub)= 54.1 kJ`..[5]

To find heat of reaction:

`N_2O_3(g) + N_2O_5(s)\rightarrow 2N_2O_4(g),\Delta H_(6,rxn) = ?`..[6]

Using Hess's law:

[5] +2 × [3] + [4] - [2] - [1] = [6]

`\Delta H_(6,rxn)=\Delta H_(5,sub)+2* \Delta H_(3,rxn)+\Delta H_(4,rxn)-\Delta H_(2,rxn)-\Delta H_(1,rxn)`

`\Delta H_(6,rxn)=54.1 kJ+(2* (-57.2 kJ))+(-114.2 kJ)-(-112.5 kJ)-(-39.0 kJ)`

`\Delta H_(6,rxn)=-23.0 kJ`

The heat of the given reaction is -23.0 kJ.