Question

Using the attached thermodynamic tables, which of the following values corresponds to the standard enthalpy of reaction ( Delta H r* S ^ ) for the reaction: C 2 H 5 OH(l)-> C 2 H 5 OH(g)^ prime ?

Answer 1

The internal energy change for the reaction is approximately -1371.5 kJ/mol.

We can use the relationship between enthalpy change (ΔH), internal energy change (ΔU), and the number of moles of gas (Δn) to find ΔU:

ΔH = ΔU + ΔnRT

where:

ΔH is the enthalpy change (given as -1366.5 kJ/mol)

ΔU is the internal energy change (what we are solving for)

Δn is the change in the number of moles of gas (3 moles of gas are produced in the reaction)

R is the ideal gas constant (8.314 J/mol K)

T is the temperature (27°C + 273.15 = 300.15 K)

Substituting the known values:

ΔU = ΔH - ΔnRT = -1366.5 kJ/mol - (3 mol) * (8.314 J/mol K) * (300.15 K)

ΔU ≈ -1371.5 kJ/mol

Question:-

The value of enthalpy change (ΔH) for the reaction C_2H_5OH ( l ) + 3O_2(g) → 2CO_2(g)+ 3 H_2O(l), at 27°C is -1366.5 kJ mol ^{−1} . The value of internal energy change for the above reaction at this temperature will be ___