How Do You Calculate the Thermodynamic Properties of Ethyl Alcohol Combustion?

[ahmedhassan72]

Homework Statement

The combustion of 1 gm of ethyl alcohol C2H5OH in abomb calorimeter evelves 29.62 KJ at 25 degree celsius.Calculate:
1-the change in internal energy of the system
2-the heat of combustion
3-delta H formation of C2H5OH

delta H Formation of CO2 gas =-393.5 KJ/mole
delta H Formation of H2O liquid=-285.58 KJ/mole
c=12 H=1 O=16

Homework Equations

The Attempt at a Solution

C2H5OH(s)+3 O2(g)-------2 CO2(g)+3 H2O(l)
1-Since a bomb calorimeter therefore the system has a constant volume therefore the heat is evolved as change in internal energy=Q(v)
=-29.62 KJ

2-The Heat of combustion is the change in internal energy of the system in that case but for 1 mole and since no of moles in the calorimeter =1/46 therefore the heat of combustion=46*-29.62=-1362.52 KJ

3-Delta H=Delta E +Delta (n)*R*T for the ordinary equation one mole ethyl alcohol
Delta H=-1362.52+(-1)(8.13)(298)(0.001)=-1364.82 KJ
Delta H=2*-393.5+3*-288.58-delta h formation of c2h5oh
there the enthalpy of formation=1434.56 KJ

 

[nate808]

1 - The change in internal energy of the system can be calculated using the equation Q = ΔU + W, where Q is the heat absorbed by the system, ΔU is the change in internal energy, and W is the work done by the system. Since the bomb calorimeter is a closed system with constant volume, W = 0. Therefore, ΔU = Q = 29.62 kJ.

2 - The heat of combustion is the amount of heat released when 1 mole of a substance undergoes complete combustion. In this case, 1 mole of ethyl alcohol (C2H5OH) is combusted, releasing 29.62 kJ of heat. Therefore, the heat of combustion is also 29.62 kJ.

3 - The enthalpy of formation (ΔHf) is the change in enthalpy when 1 mole of a compound is formed from its constituent elements in their standard states. Using the given values for the enthalpies of formation of CO2 and H2O, we can calculate the enthalpy of formation of C2H5OH as follows:

C2H5OH + 3/2 O2 → 2 CO2 + 3 H2O

ΔHf(C2H5OH) = 2ΔHf(CO2) + 3ΔHf(H2O) - ΔHf(C2H5OH)
= 2(-393.5 kJ/mol) + 3(-285.58 kJ/mol) - (-1364.82 kJ/mol)
= -787 kJ/mol - 856.74 kJ/mol + 1364.82 kJ/mol
= -1364.92 kJ/mol

Therefore, the enthalpy of formation of ethyl alcohol is -1364.92 kJ/mol.

 

[Blueshift5]

/mole

I would like to commend the student for their attempt at solving the given problem. However, I would like to offer some clarifications and corrections to their solution.

1. The change in internal energy of the system should be calculated using the formula ΔU = Q + W, where Q is the heat and W is the work done by the system. In this case, since the system is a bomb calorimeter, there is no work done (W=0). Therefore, the change in internal energy is simply equal to the heat evolved, which is 29.62 kJ.

2. The heat of combustion should be calculated using the formula ΔH = ΔU + PΔV, where P is the pressure and ΔV is the change in volume. In this case, since the pressure is constant and the volume is negligible, ΔH is also equal to the heat evolved, which is 29.62 kJ.

3. For the calculation of ΔH formation, the correct formula is ΔH = ΔU + PΔV + nRT, where n is the number of moles and R is the gas constant. In this case, since the pressure is constant and the volume is negligible, ΔH is equal to ΔU + nRT. Therefore, the correct value for ΔH formation of C2H5OH is -1364.82 kJ.

Overall, the student's approach to solving the problem was correct, but the formulas used were not entirely accurate. it is important to always double-check and verify the equations and values used in calculations to ensure accurate results.