Question

Given that 2NH3 + N2 + 3H2 ; Delta H = 46 k.cal
From the above reaction, heat of formation of
ammonia is
46 k.cal
-46 k.cal
-23 k.cal
23 k.cal

Answer 1

Explanation:

ANSWER

The enthalpy of formation of ammonia is −46.0 kJ mol

−1

.

1/2 N

2

(g)+3/2 H

2

(g)→NH

3

(g) ΔH=−46.0kJ/mol

Multiply above equation with 2.

N

2

(g)+3H

2

(g)→2NH

3

(g) ΔH=−46.0×2=−92.0kJ/mol

Reverse above equation

2NH

3

(g)→N

2

(g)+3H

2

(g) ΔH=+92.0kJ/mol

The enthalpy change for the reaction 2NH

3

(g)→N

2

(g)+3H

2

(g) is 92.0 KJ mol

−1