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Ammonia, NH3 (Hf =  –46.2 kJ), reacts with oxygen to produce water (Hf = –241.8 kJ) and nitric oxide, NO (Hf = 91.3 kJ), in the following reaction: What is the enthalpy change for this reaction?  Use . –900.8 kJ –104.6 kJ 104.6 kJ 900.8 kJ

Respuesta :

Given enthalpy of formation (Hf) data:

Hf(NH3) = -46.2 KJ

Hf(H2O) = -241.8 kJ

Hf(NO) = 91.3 kJ

To determine:

Enthalpy change for the following reaction-

4NH3 + 7O2 → 4NO + 6H2O

Explanation:

The enthalpy change for this reaction is given as:

ΔH = ∑n*Hf(products) - ∑n*Hf(reactants)

     = [4(91.3) + 6(-241.8)] -[4(-46.2) + 7(0)] = -900.8 kJ

Ans: (a)

Reaction enthalpy = -900.8 kJ

Answer : The enthalpy change for this reaction is, -900.8 KJ

Solution :

The balanced chemical reaction is,

[tex]4NH_3(l)+5O_2(g)\rightarrow 4NO(l)+6H_2O(l)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{H_2O}\times \Delta H_{H_2O})+(n_{NO}\times \Delta H_{NO})]-[(n_{NH_3}\times \Delta H_{NH_3})+(n_{O_2}\times \Delta H_{O_2})][/tex]

where,

n = number of moles

Now put all the given values in this expression, we get

[tex]\Delta H=[(6\times -241.8)+(4\times 91.3)]-[(4\times -46.2)+(5\times 0)]\\\\\Delta H=-900.8KJ[/tex]

Therefore, the enthalpy change for this reaction is, -900.8 KJ