Given that for the reaction. N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H = -92kJ. What is...

WAEC 1989

Given that for the reaction. N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H = -92kJ. What is the enthalpy of formation of ammonia from its elements?

A. -46KJ mol ^-1
B. -184KJmol ^-1
C. -92KJmol ^-1
D. 184KJmol ^-1
E. +46KJmol ^-1

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Correct Answer: Option A

Explanation

In the reaction the enthalpy of formation 2 mole of NH3 is - 92kJ. Therefore the standard \(\Delta H\) of formation = \(\frac{-92kJ}{2} = -46KJmol^{-1}\).

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Given that for the reaction. N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H = -92kJ. What is...

Correct Answer: Option A

Explanation

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