Use the information below to answer this question C(s) + O2(g) → CO2(g) ∆H = −393.5 kJ mol−1 H2(g) + O2(g) → H2O(l) ∆H = −285.8 kJ mol−1 3C(s) + 4H2(g) → C3H8(g) ∆H = −104.0 kJ mol−1 4C(s) + 5H2(g) → C4H10(g) ∆H = −125.2 kJ mol−1 The value in kJ mol−1 for the enthalpy of combustion of propane is
First, construct a balance equation for the combustion of propane. Then the 3rd equation given in the question is the formation of propane. So 1 mol of propane is formed by -104 kJ mol-1. Then, the 1st equation given in question is the formation of CO2. 3 mol of CO2 is formed so 3(-393.5). 2nd equation shows that 1 mol of H2O is formed but based on balanced equation that I constructed, 4 mol of H2O is formed so 4(-285.5). The arrows I drew was for the mathematics equation. Going to same direction of the product of combustion of propane FROM the reactant is -104 and x so I put this on the left side. Then, the remaining one is on the right side.