Question

What is ΔU for the formation of 1 mole of CO at 1 atm and 25∘C ? C(graphite) ​+1/2O₂​( g)→CO(g)​ΔH=−110.5 kJ/ mol Note: R=8.314 J/K⋅mol

Answer 1

To find the change in internal energy (ΔU) for the formation of 1 mole of CO at 1 atm and 25°C, one can use the enthalpy change (ΔH) for the reaction C(graphite) + 1/2 O₂(g) → CO(g), which is –110.5 kJ/mol, and account for the change in moles of gas and temperature to calculate ΔU.

Step-by-step explanation:

The change in internal energy (ΔU) for the formation of 1 mole of CO at 1 atm and 25°C can be related to the change in enthalpy (ΔH) for the reaction using the equation ΔU = ΔH - ΔnRT, where Δn is the change in the number of moles of gas and R is the universal gas constant. For the formation of carbon monoxide (CO) from graphite and oxygen, the reaction is C(graphite) + 1/2 O₂(g) → CO(g) with a ΔH of –110.5 kJ/mol. The change in the number of moles of gas is 1 - (1 + 1/2) = -1/2 moles. Using R = 8.314 J/K⋅mol and T = 298.15 K (25°C in Kelvin), we can then calculate ΔU.

ΔU = –110.5 kJ/mol - (-1/2 moles ⋅ 8.314 J/K⋅mol ⋅ 298.15 K)
ΔU = -110.5 kJ/mol - (-1/2 ⋅ 8.314 ⋅ 298.15 J)

After converting kJ to J to ensure consistent units (1 kJ = 1000 J), and performing the calculations, ΔU can be obtained as the final answer.

Answer 2

The change in internal energy (ΔU) for the formation of 1 mole of CO at 1 atm and 25°C is −111.737 kJ/mol. This is determined by adjusting the given enthalpy change (ΔH) for the work done by gases expanding or contracting under constant pressure.

Step-by-step explanation:

The student is asking for the change in internal energy (ΔU) for the formation of 1 mole of carbon monoxide (CO) at standard conditions of 1 atm and 25°C. The provided reaction is:

C(graphite) + ½O₂(g) → CO(g) with ΔH = −110.5 kJ/mol.

The change in internal energy (ΔU) can be estimated from the enthalpy change (ΔH) using the relation:

ΔU = ΔH - (nRT)

where n is the change in moles of gas during the reaction, R is the ideal gas constant, and T is the temperature in Kelvin. Since ΔH given is for a 1 mole formation of CO, we need to account for the ½ mole change in gas from reactants to products. At 25°C (which is 298.15 K), using R (8.314 J/K∙mol), we get:

ΔU ≈ −110.5 kJ/mol - (½ mol × 8.314 J/K∙mol × 298.15 K) = −110.5 kJ/mol - (1.237 kJ/mol) = −111.737 kJ/mol.

This calculation assumes that the pressure and temperature remain constant, and uses the provided value of R and the conversion of temperature to Kelvin.