Question

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

Answer 1

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

Solution :

The balanced chemical reaction is,

`4NH_3(l)+5O_2(g)\rightarrow 4NO(l)+6H_2O(l)`

The expression for enthalpy change is,

`\Delta H=\sum [n* \Delta H_f(product)]-\sum [n* \Delta H_f(reactant)]`

`\Delta H=[(n_(H_2O)* \Delta H_(H_2O))+(n_(NO)* \Delta H_(NO))]-[(n_(NH_3)* \Delta H_(NH_3))+(n_(O_2)* \Delta H_(O_2))]`

where,

n = number of moles

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

`\Delta H=[(6* -241.8)+(4* 91.3)]-[(4* -46.2)+(5* 0)]\\\\\Delta H=-900.8KJ`

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

Answer 2

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

Step-by-step 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