Question

For the reaction at 298 K, 1 2 N2 (g) + 3 2 H2 (g) → NH3 (g) ; Δ H = - 46 kJ(a) What is the value of Δng?
(b) Calculate the value of Δ U at 298 K?

Answer 1

Answer:

Δn = - 2

ΔU  = -41.044 KJ

Explanation:

Considering the above equation;

1/2 N2 (g) + 3/2 H2 (g) → NH3 (g)

Δn = 1 – ½ -3/2  

Δn = (2 – 1 -3)/2

Δn = - 2

Standar equation for calculating enthalpies and energy changes

ΔH =    ΔU  + ΔnRT

-46000 = ΔU  + (-2)(8.314)(298)

-46000 = ΔU  - 4955.144

ΔU  = -41044.856

ΔU  = -41.044 KJ