Question

the the enthalpy change (∆H) for the reaction
N2 (g) + 3H2 (g) ------------>2NH3 (g)
is 92.38kJ at 298k . what is ∆E at 298k.??​

Answer 1

Answer:

∆H and ∆E are related as

∆H = ∆E +∆ngRT............(1).

for the reaction

N2(g) + 3H2(g) -------> 2NH3(g)

∆ng = 2 - (1 + 3) = -2

∆H = -92.38 kJ

R = 8. 314 x 10^-13 kJ k^-1 mol^-1

T = 398 k

So....

From (1).....

-92.38 = ∆E +(-2) x 8. 314 x 10^-3 x 298

-92.38 = ∆E - 4.96

∆E = -92.38 + 4.96

= - 87.42 kJ

Answer 2

Explanation:

we know that ∆H = ∆E + ∆ngRT

∆ng= -2 (by solution)

∆H.= -92.38kj (given)

R= 8.314x10^-13kj

From all this equations,

∆H= ∆E + ∆ngRT (applying this formula)

∆E = 4.96+(-92.38)

= 87.42kj

Hope this helps you