Standard heat of formation of ammonia is - x kJ/mol.
The heat of reaction when 28 gm of N2 (g) reacted with 6 g of H2 (g) at S.T.P
(1) -x kJ
(2) -2x kJ
(3) x kJ
(4) 2x kJ
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Answers
Answer :
option (2) -2x kJ
Explanation :
Standard heat of formation :
It is the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements (in normal physical state)
Given,
Standard heat of formation of ammonia is -x kJ/mol
(negative sign represents heat is evolved)
we have to find the heat of reaction when 28 gm of N₂ (g) reacted with 6 g of H₂ (g) at S.T.P
For that ; the number of moles of ammonia formed at the given condition is to be found.
The balanced equation for the formation of ammonia is
N₂(g) + 3H₂(g) ⟶ 2NH₃(g)
⇒ 28 g of nitrogen (N₂)
number of moles = 28/28 = 1
⇒ 6 g of hydrogen (H₂)
number of moles = 6/2 = 3
From the above chemical equation,
1 mole of nitrogen reacted with 3 moles of hydrogen forming 2 moles of ammonia.
So, 28 g of nitrogen reacts with 6 g of hydrogen to form 2 moles of ammonia.
The heat of reaction for one mole of ammonia = -x kJ
The heat of reaction when two moles of ammonia are formed
= 2(-x) kJ
= -2x kJ
Therefore, The heat of reaction when 28 gm of N₂ (g) reacted with 6 g of H₂ (g) at S.T.P is -2x kJ
option (2) ⇒ -2x Kj