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Question

The enthalpy change for a reaction N2(g)+3H2(g)2NH3(g) is -92.2 kJ/mol. Calculate the enthalpy of formation of ammonia.

A
-92.2 kJ/mol
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B
-46.1 kJ/mol
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C
46.1 kJ/mol
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D
-92.2 kJ/mol
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Solution

The correct option is B -46.1 kJ/mol
Given, N2(g)+3H2(g)2NH3(g) ΔHr=92.2 kJ

Here 2 moles of ammonia are forming but according to definition of enthalpy of formation we know it is for formation of 1 mole of substance. i.e for formation of 1 mole of ammonia

dividing whole equation by 2 we get
12N2(g)+32H2(g)NH3(g)
ΔHf(NH3)=92.22=46.1 kJ
hence, heat of formation of ammonia is 46.1 kJ/mol

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