Ammonia reacts with oxygen to form nitrogen dioxide and steam: 4NH 3 (g) +7O 2 (

Question

Ammonia reacts with oxygen to form nitrogen dioxide and steam:

4NH3(g) +7O2(g) --> 4NO2(g) + 6H2O(g)

Given the following standard enthalpies of formation (given below in kJ.mol), calculate the enthalpy of reaction.

NH3(g) = -45.90

NO2 (g) = +33.1

Ammonia reacts with oxygen to form nitrogen dioxide and steam, as follows: 4NH3(g) +702(g)4NO2(g) + 6H20(g) Given the following standard enthalpies of formation (given below in kJ/mol), calculate the enthalpy 4. reaction. NH3(g) -45.90 NO2(g) +33.1 H20(g) -241.83

Explanation / Answer

The enthalpy of the reaction can be calculated by subtracting the total enthalpies of formation of reactants from that of products.

The enthalpy of products :

= 4(33.1) + 6(-241.83)

= -1318.58 KJ

Enthalpy of reactants :

=4(-45.90) + 7(0)

= -183.6 KJ

So the enthalpy of the reaction

= -1318.58 - (-183.6)

= 1134.97 KJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.