Question

1) Consider the combustion of coal (C₇₀H₄₁O₆N) with a heat of formation of -300kJ/mol in 20 mol% excess air with main combustion products of CO₂ , H₂O, and NO₂ .

a) Write the balanced form of this reaction

b) Calculate the standard heat of reaction

c) Calculate the heat of reaction at 1600⁰ C

Answer 1

a) Balanced equation:
4 C₇₀H₄₁O₆N + 313 O₂ = 280 CO₂ + 82 H₂O + 4 NO₂

_b) The standard heat of reaction :

To calculate the standard enthalpy of reaction the standard enthalpy of formation must be utilized. Another, more detailed, form of the standard enthalpy of reaction includes the use of the standard enthalpy of formation ΔHºf:

ΔH^⊖=∑ΔvpΔH^⊖f(products)−∑ΔvrΔH^⊖f(reactants)

with

• vp= stoichiometric coefficient of the product from the balanced reaction
• vr= stoichiometric coefficient of the reactants from the balanced reaction
• ΔHºf= standard enthalpy of formation for the reactants or the products

ΔH^⊖ = -366X300 + 317X300 = -14700 Joules

Exothermic, since a negative value indicates that the system Lost heat.

A negative value for ΔHº represents a removal of energy from the reaction (and into the surroundings) and so the reaction is exothermic.

c) Exactly I dont about this problem, but by using below formulae we can calculate ( not sufficient information )

ΔH⁰(T)=ΔH⁰T1+∫ ^T1ΔC⁰_p(T′)dT

ΔH⁰(16000)= -14700 +