Enthalpy of formation of NH₃ is - 46 kJ/kg mole. The enthalpy change for the gaseous reaction, 2NH₃ → N₂ + 3H₂, is equal to __________ kJ/kg. mole.

ANS:B - 92

To find the enthalpy change for the gaseous reaction 2𝑁𝐻3→𝑁2+3𝐻22NH3​→N2​+3H2​, we can use Hess's law, which states that the total enthalpy change for a reaction is the same, regardless of the route taken. Given the enthalpy of formation of NH3 is -46 kJ/mol, we need to use this information along with the enthalpies of formation for N2 and H2, which are zero under standard conditions. The enthalpy change for the reaction can be calculated as follows: Δ𝐻reaction=∑Δ𝐻products−∑Δ𝐻reactantsΔHreaction​=∑ΔHproducts​−∑ΔHreactants​ Δ𝐻reaction=(0 kJ/mol+0 kJ/mol)−(−46 kJ/mol)ΔHreaction​=(0kJ/mol+0kJ/mol)−(−46kJ/mol) Δ𝐻reaction=46 kJ/molΔHreaction​=46kJ/mol However, the given enthalpy of formation is provided per kg mole (kg mol). So, to find the enthalpy change per kg mole, the value should remain the same. Thus, the enthalpy change for the gaseous reaction 2𝑁𝐻3→𝑁2+3𝐻22NH3​→N2​+3H2​ is 46 kJ/kg mole.