Question

Determine the difference between \(\Delta H\) and \(\Delta E\) at \(25^{\circ} \mathrm{C}\) for $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$

Short answer

Answer: The difference between ΔH and ΔE for the given reaction at 25°C is -2473.7861 J/mol.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the given reaction is already provided: $$ 2 \mathrm{CO}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$

Calculate the change in moles of gas

We'll find the change in the number of moles of gaseous molecules between the reactants and products. Change in moles of gases = moles of gaseous products - moles of gaseous reactants $$ \Delta n_g = (\mathrm{2~moles~CO_2})-(\mathrm{2~moles~CO}+\mathrm{1~mol~O_2}) $$ $$ \Delta n_g = 2 - 2 -1 = -1 $$

Apply the ideal-gas law

Assuming the ideal-gas law, we will use the relationship: $$ \Delta(PV) = \Delta n_g RT $$ Here, Δn_g is the change in moles of gas, R is the gas constant (8.314 J/mol K), and T is the temperature in the Kelvin scale (T = 25°C +273.15 = 298.15 K). Now we will calculate Δ(PV): $$ \Delta(PV) = (-1)(8.314\,\text{J/mol K})(298.15\,\text{K}) $$ $$ \Delta(PV) = -2473.7861\,\text{J/mol} $$

Determine the change in enthalpy (ΔH) and the change in internal energy (ΔE)

We will use the relationship between ΔH and ΔE: $$ \Delta H = \Delta E + \Delta(PV) $$ In order to find the difference between them, we need to rearrange the equation as follows: $$ \Delta H - \Delta E = \Delta(PV) $$ Now, we can substitute the value of Δ(PV) obtained in step 3: $$ \Delta H - \Delta E = -2473.7861\,\text{J/mol} $$

Result

The difference between ΔH and ΔE for the given reaction at 25°C is -2473.7861 J/mol.