1. At 25 degees C, (triange)H(little o) for the combustion of liquid octane, C8H18(l) to form CO2(g) and H2O(l) is -5471kJ/mol. Calculate (triangle)E(little o) for this reaction at 25 degrees C.
2. The molar heat of vaporization of benzene, C6H6, at 1 atm is 30.8 kJ/mol. The boiling point of benzene at 1 atm is 78.3 degree C. Calculate the entropy change when 11.5g of benzene boils at the temperature.
3. The heat capacity of a bomb calorimeter, is 87.5kJ/K (this value is for the total heat capacity including that of the water jacket around the reaction chamber). If 67.2g of methane(g) is combusted under such reaction conditions, what will be the increase in temperature on the calorimeter? What if (triange)E combustion for CH4(g) is -885.4 kJ/mol?
4. The standard free energies of the formation of SO2(g) is -300.4 kJ/mol and SO3(g) is -370.4 kJ/mol. Calculate the value of the equilibrium constant at 25 degrees C for the reaction:
2SO2(g) + O2 ---------- 2SO3(g)
Please find the attached solution.
1.Given : At 298 K ΔH0 for the combustion of liquid octane is -5471 KJ /mol
We need to calculate ΔU0 for the same.
We know that, ΔH0 = ΔH0 + ΔngRT
where Δng is the difference between the no. of gaseous products and the
gaseous reactants. The combustion of octane can be written as :
2 C8H18 (l) + 18O2 (g) ==== 16 CO2 (g) + 18 H2O (l)
thus we see that Δng = -9 and therefore : ...
This solution involves various calculation of thermodynamics: heat of combustion, entropy change, and equilibrium constant.