Question

Propane has the structure CH3-CH2-CH3. Calculate the enthalpy change, ∆rH° for the reaction:

C3H8(g) + 5O2 (g)  →→3CO2 (g) + 4H2O (g)

The average bond enthalpies of various bonds are

C-C   →→   347 KJ/mol

C-H   →→   414 KJ/mol

O=O  →→   498 KJ/mol

C=O  →→   741 KJ/mol

O-H    →→  464 KJ/mol​

Answer 1

Answer:

-1234 kJ/mol

Explanation:

The negative sign means that the reaction is exothermic

EXTRA STUFF→

Standard enthalpy change (∆H°) is the enthalpy change in standard conditions

i.e. when temperature is 298 k

It can be find using HESS'S LAW of constant heat summation

Answer 2

Explanation:

Answer is -1234 kj/mol . you can calculate ii by hess law. also negative sign is for exothermic reactions