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Calculate the change in enthalpy for the following reaction:
4NH3 (g) + 7O2 (g) --> 4 NO2 (g) + 6 H20 (l)
AH NH3= -46
AH4NO2= +34
AH+H200=-286
AH reaction= ____ kJ/mol
Is this reaction endothermic or exothermic?

Sagot :

Dsdrajlinidn

Answer:

-1396 kJ

Explanation:

Let's consider the following balanced equation.

4 NH₃(g) + 7 O₂(g) ⇒ 4 NO₂(g) + 6 H₂O(l)

We can calculate the standard change in enthalpy for the reaction (ΔHr) using the following expression.

ΔHr = ∑ np × ΔHf(p) - ∑ nr × ΔHf(r)

where,

  • n: moles
  • ΔHf: standard enthalpies of formation
  • p: products
  • r: reactants

ΔHr = 4 mol × ΔHf(NO₂(g)) + 6 mol × ΔHf(H₂O(l)) - 4 mol × ΔHf(NH₃(g)) - 7 mol × ΔHf(O₂(g))

ΔHr = 4 mol × (34 kJ/mol) + 6 mol × (-286 kJ7mol) - 4 mol × (-46 kJ/mol) - 7 mol × (0 kJ/mol) = -1396 kJ

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