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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

Respuesta :

Answer:

The heat of the given reaction is -23.0 kJ.

Explanation:

We are given with ;

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

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

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

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

[tex]N_2O_5(s)\rightarrow N_2O_5(g) ,\Delta H_{5,sub}= 54.1 kJ[/tex]..[5]

To find heat of reaction:

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

Using Hess's law:

[5] +2 × [3] + [4] - [2] - [1] = [6]

[tex]\Delta H_{6,rxn}=\Delta H_{5,sub}+2\times \Delta H_{3,rxn}+\Delta H_{4,rxn}-\Delta H_{2,rxn}-\Delta H_{1,rxn}[/tex]

[tex]\Delta H_{6,rxn}=54.1 kJ+(2\times (-57.2 kJ))+(-114.2 kJ)-(-112.5 kJ)-(-39.0 kJ)[/tex]

[tex]\Delta H_{6,rxn}=-23.0 kJ[/tex]

The heat of the given reaction is -23.0 kJ.

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