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:

smooth muscle is composed of involuntary muscles found within the walls of organs and structures such as the esophagus, stomach, intestines, and blood vessels. those muscles push materials like food or blood through organs. unlike skeletal muscle, smooth muscle can never be under your control.