Question

Gaseous Nitrous oxide decomposes at 298 K to form Nitrogen gas and Oxygen gas. The \(\Delta H \text{ for the reaction at 1.0 atm pressure and 298 K is } \)-163.15 \(\text{ kJ}\). Calculate Internal energy change for the decomposition of 100 g of Nitrous oxide gas under the same conditions of temperature and pressure.

• A. -166 kJ
• B. -188.2 kJ
• C. -230.3 kJ
• D. -376.43 kJ

Hint:

When calculating changes in internal energy for reactions, make sure to account for the change in moles of gas (\(\Delta n_g\)) and use the correct temperature and pressure values.

Answer 1

The Correct Option is B

We are given that the enthalpy change (\(\Delta H\)) for the reaction is \(-163.15 \, \text{kJ}\), and we need to calculate the internal energy change (\(\Delta U\)) for the decomposition of 100 g of Nitrous oxide. The relation between \(\Delta H\) and \(\Delta U\) is given by the equation:
\[ \Delta H = \Delta U + \Delta n_g RT \] Where: - \(\Delta H = -163.15 \, \text{kJ}\), - \(\Delta n_g = 1 \, \text{mol}\) (the change in moles of gas), - \(R = 8.314 \, \text{J/mol·K}\), - \(T = 298 \, \text{K}\). Now, we calculate the internal energy change: \[ \Delta U = \Delta H - \Delta n_g RT \] \[ \Delta U = -163.15 \, \text{kJ} - (1 \times 8.314 \times 298) \times \frac{1}{1000} \] \[ \Delta U = -163.15 \, \text{kJ} - 2.478 \, \text{kJ} = -165.628 \, \text{kJ} \] Next, we calculate the total internal energy change for 100 g of Nitrous oxide: \[ \text{Moles of N}_2\text{O} = \frac{100}{44} = 2.27 \, \text{mol} \] \[ \Delta U (\text{total}) = \Delta U (\text{per mole}) \times \text{moles of N}_2\text{O} \] \[ \Delta U (\text{total}) = -165.628 \times 2.27 = -188.2 \, \text{kJ} \] Final Answer:
\boxed{-188.2 \, \text{kJ}}