Calculate the enthalpy change for reaction in kjmol-1

I got

−

902 kJ

for the reaction as-written. How would you rewrite this into

kJ/mol NH

3

?

The standard molar enthalpy of formation is for the formation of

1 mol

of product from its elements as they exist in nature at

25

∘

C

and

1 atm

. For example...

1

2

N

2

(

g

)

+

3

2

H

2

(

g

)

→

NH

3

(

g

)

,

Δ

H

∘

f

(

NH

3

(

g

)

)

=

−

45.9 kJ/mol

But that also means formation of elements in their standard states must yield zero enthalpy change:

1

2

O

2

(

g

)

→

1

2

O

2

(

g

)

,

Δ

H

∘

f

(

O

2

(

g

)

)

=

0

That means we can simply derive from Hess's law and that fact to get:

Δ

H

∘

r

x

n

=

∑

P

n

P

Δ

H

∘

f

,

P

−

∑

R

n

R

Δ

H

∘

f

,

R

where

n

indicates the mols of product

P

or reactant

R

. Here we actually have just summed many formation reactions together that we know have

Δ

H

∘

r

x

n

=

Δ

H

∘

f

.

Here we have:

Δ

H

∘

r

x

n

=

[

4 mols NO

(

g

)

⋅

Δ

H

∘

f

,

N

O

(

g

)

+

6 mols H

2

O

(

g

)

⋅

Δ

H

∘

f

,

H

2

O

(

g

)

]

−

[

4 mols NH

3

(

g

)

⋅

Δ

H

∘

f

,

N

H

3

(

g

)

+

0 kJ/mol

]

where we have set a zero contribution

O

2

(

g

)

right off the bat.

=

[

4 mols NO

(

g

)

⋅

91.3 kJ/mol

+

6 mols H

2

O

(

g

)

⋅

−

241.8 kJ/mol

]

−

[

4 mols NH

3

(

g

)

⋅

−

45.9 kJ/mol

+

0 kJ/mol

]

=

[

−

1085.6 kJ

]

−

[

−

183.6 kJ

]

=

−

902 kJ

for the reaction as-written.

Explanation:

dynamic equilibrium is the answer.