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

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,

The expression for enthalpy change is,

where,

n = number of moles

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

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

the answer is true.

2b2t

explanation:

2b2t