Question

Calculate the standard enthalpy change for the reaction 

2C8H18(I) + 17O2(g) → 16CO(g)+ 18H2O(1) 

Given: 2C8H18(I) + 25O2(g) → 16CO2(g)+18H2O(I)  ΔH° = -11,020 kJ/mol

2CO(g) + O 2(g) → 2CO2(g)  ΔH° = -566.0 kJ/mol 

• -10.450 kJ/mol 

• -6,492 kJ/mol 

• 6,492 kJ/mol 

• 10,450 kJ/mol 

• 15,550 kJ/mol

Answer 1

Answer 2

Lets number the reaction as 0, 1, 2 from top to bottom

required reaction should be written in terms of other reaction

This is Hess Law

required reaction can be written as:

reaction 0 = +1 * (reaction 1) -8 * (reaction 2)

So, ΔHo rxn for required reaction will be:

ΔHo rxn = +1 * ΔHo rxn(reaction 1) -8 * ΔHo rxn(reaction 2)

= +1 * (-11020.0) -8 * (-566.0)

= -6492 KJ/mol

Answer: -6492 KJ/mol

Answer 3

Lets number the reaction as 0, 1, 2 from top to bottom

required reaction should be written in terms of other reaction

This is Hess Law

required reaction can be written as:

reaction 0 = +1 * (reaction 1) -8 * (reaction 2)

So, ΔHo rxn for required reaction will be:

ΔHo rxn = +1 * ΔHo rxn(reaction 1) -8 * ΔHo rxn(reaction 2)

= +1 * (-11020.0) -8 * (-566.0)

= -6492 KJ

Answer: -6492 KJ