Question

For the reaction CH\(_3\)COOH(l) + 2O\(_2\)(g) \(\rightarrow\) 2CO\(_2\)(g) + 2H\(_2\)O(l) at 25°C and 1 atm pressure, \(\Delta H = -874\) kJ. Then the change in internal energy (\(\Delta E\)) is

• A. -874 kJ
• B. -871.53 kJ
• C. -876.47 kJ
• D. +874 kJ

Hint:

When \(\Delta n_g = 0\), \(\Delta H = \Delta E\).

Answer 1

The Correct Option is A

To find the change in internal energy, \(\Delta E\), for the reaction given, we use the relationship between enthalpy change (\(\Delta H\)) and internal energy change (\(\Delta E\)). The equation relating these two terms is:

\(\Delta H = \Delta E + \Delta n_gRT\)

where:

• \(\Delta H\) is the change in enthalpy
• \(\Delta E\) is the change in internal energy
• \(\Delta n_g\) is the change in the number of moles of gases
• \(R\) is the universal gas constant, which is \(8.314 \, \text{J/mol} \cdot \text{K}\)
• \(T\) is the temperature in Kelvin (25°C = 298 K)

First, let's calculate \(\Delta n_g\), the change in moles of gaseous substances:

• Moles of gas on the reactant side: 2 (from \(O_2\))
• Moles of gas on the product side: 2 (from \(CO_2\))
• Therefore, \(\Delta n_g = 2 - 2 = 0\)

Substituting into the equation:

\(\Delta H = \Delta E + 0 \times RT\)

\(\Delta H = \Delta E\)

This simplifies to \(\Delta E = \Delta H\), since any term multiplied by zero is zero.

Given that \(\Delta H = -874 \, \text{kJ}\), we find \(\Delta E = -874 \, \text{kJ}\). Thus, the change in internal energy is:

-874 kJ

Therefore, the correct answer is -874 kJ.