Conversion of Enthelpy (Combustion)

[skaboy607]

Homework Statement

The enthalpy of combustion ∆Ho at t = 25 0C of Benzene (C6H6) is –3169540 kJ/kmol with the H2O in the vapour phase. Calculate the internal energy of combustion ∆Uo in kJ/kg. Calculate also ∆Ho for H2O in the liquid state.

Homework Equations

ΔH° = ΣΔH°f(products) - ΣΔH°f(reactants)

The Attempt at a Solution

The temperature of the reactants and products is the same so I know that I can just use the enthalpy of combustion given and then subtract the h(fg) value from steam tables to give me my answer for the liquid state. I am struggling on how to get the internal energy of combustion.

Thanks for your help.

 

[KataKoniK]

To calculate the internal energy of combustion, you can use the equation:

ΔU° = ΔH° - ΔnRT

where ΔH° is the enthalpy of combustion, Δn is the change in moles of gas (in this case, it would be -1 since the reactant is C6H6 and the product is H2O in the gas phase), R is the gas constant (8.314 J/mol K), and T is the temperature in Kelvin.

So, for the given reaction, the internal energy of combustion would be:

ΔU° = -3169540 kJ/kmol - (-1 mol)(8.314 J/mol K)(298 K) = -3169540 kJ/kmol + 2469 kJ/kmol = -3167071 kJ/kmol

To calculate the enthalpy of combustion for H2O in the liquid state, you can use the equation:

ΔH° = ΔH°vap + ΔH°f

where ΔH°vap is the enthalpy of vaporization and ΔH°f is the enthalpy of formation. You can find these values in steam tables or other reliable sources.

I hope this helps. Let me know if you have any further questions. Good luck with your calculations!

 

[Mene]

I would approach this problem by first reviewing the equations and principles related to enthalpy and internal energy of combustion. Enthalpy is a measure of the heat energy released or absorbed during a chemical reaction, while internal energy is a measure of the total energy contained within a system. In the case of combustion, both enthalpy and internal energy are important in understanding the energy changes that occur.

To calculate the internal energy of combustion, we can use the equation ∆U = ∆H - ∆PV, where ∆H is the change in enthalpy, ∆P is the change in pressure, and ∆V is the change in volume. Since this equation includes pressure and volume, it is important to specify the conditions under which the enthalpy of combustion was measured. In this case, the enthalpy of combustion was measured at a temperature of 25°C and with the H2O in the vapor phase. Therefore, we can assume that the pressure and volume are constant and can be ignored in the calculation of internal energy.

Using the given enthalpy of combustion for benzene, we can calculate the internal energy of combustion using the equation ∆U = ∆H = -3169540 kJ/kmol. This value represents the energy released during the combustion of 1 kmol of benzene.

To calculate the enthalpy of H2O in the liquid state, we can use the equation ΔH° = ΔH°f(liquid) - ΔH°f(vapor), where ΔH°f(liquid) is the enthalpy of formation of liquid water and ΔH°f(vapor) is the enthalpy of formation of water vapor. By looking up these values in a table of thermodynamic properties, we can calculate the enthalpy of H2O in the liquid state as -285.8 kJ/kg. This value represents the amount of energy released when 1 kg of water vapor condenses to form liquid water.

In summary, the internal energy of combustion of benzene is -3169540 kJ/kmol, and the enthalpy of H2O in the liquid state is -285.8 kJ/kg. These values can be used to further analyze the energy changes that occur during the combustion of benzene and the formation of liquid water.