Question

The enthalpy of formation of ammonia is
--46.0 kJ moll The enthalpy change for the
reaction 2NH2(g) → N.(g) + 3H2(g) is :
(1) 46.0 kJ mol-
(2) 92.0 kJ mol-1
(3) - 23.0 kJ mol-1 (4) - 92.0 kJ mol-1​

Answer 1

$\huge\underline\green{\sf Answer}$

(4) -92.0KJ mol-1

$\huge\underline\green{\sf Solution}$

Given Reaction :-

$\large{2NH_{3}->N_{2}+3H_{2}}$

∆$\large{H_{reaction}}$=(2×∆H of $\large{NH_{3}}$)-(∆H of + 3∆H of $\large{H_{2}}$)

∆$\large{H_{reaction}}$=2×(-46)-0

✭ ∆H = -92 kJ for N2

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