To determine in which chemical reaction the change in internal energy (\(\Delta E\)) is equal to the change in enthalpy (\(\Delta H\)), we must understand the relationship between these two quantities. In a chemical reaction, the relationship is given by the formula:
\(\Delta H = \Delta E + \Delta n_gRT\)
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 gas.
- \(R\) is the universal gas constant.
- \(T\) is the temperature in Kelvin.
For \(\Delta H = \Delta E\), the term \(\Delta n_gRT\) must be zero, which implies that \(\Delta n_g = 0\). This means there is no change in the number of moles of gas when moving from reactants to products.
Let's evaluate the options:
- \(N_2O_4(g) \rightleftharpoons 2NO_2(g)\)
- Reactant moles = 1
- Product moles = 2
- \(\Delta n_g = 2 - 1 = 1\). Therefore, \(\Delta H \neq \Delta E\).
- \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\)
- Reactant moles = 2 + 1 = 3
- Product moles = 2
- \(\Delta n_g = 2 - 3 = -1\). Therefore, \(\Delta H \neq \Delta E\).
- \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\)
- Reactant moles = 1 + 1 = 2
- Product moles = 2
- \(\Delta n_g = 2 - 2 = 0\). Therefore, \(\Delta H = \Delta E\).
- \(H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(l)\)
- Reactant moles = 1 + 0.5 = 1.5
- Product moles = 0 (as water is liquid, not gas)
- \(\Delta n_g = 0 - 1.5 = -1.5\). Therefore, \(\Delta H \neq \Delta E\).
The correct answer is: \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\), because in this reaction, there is no change in the number of moles of gas.