Question

N2(g) +3N2(g) _2NH3(g) +92kj, H2 मिला ने पर​

Answer 1

Answer:

For the given reaction,

N

2.

(g)+3H

2

(g)⟶2NH

3

(g)

Enthalpy of reaction (Δ

r

H

0

)= –92.4 kJ mol

–1

N

2

(g)+H

2

(g)⟶NH

3

(g)

Hence, standard enthalpy of NH

3

is equal to the

2

1

Δ

r

H

0

As, Δ

r

H

0

= –92.4 kJ mol

–1

∴ Standard enthalpy of NH

3

= 2

−92.4

⇒Standard enthalpy of NH

3

= −46.2 kJ mol

−1