The enthalpy of formation of NH3 is – 46. The enthalpy change for the reaction: 2NH3 → N2 + 3H2 is ?46 KJ92 KJ–23 KJ –92 KJ
Question
The enthalpy of formation of NH3 is – 46. The enthalpy change for the reaction: 2NH3 → N2 + 3H2 is ?46 KJ92 KJ–23 KJ –92 KJ
Solution
To find the enthalpy change for the reaction 2NH3 → N2 + 3H2, we can use the enthalpy of formation values for NH3, N2, and H2.
Given that the enthalpy of formation of NH3 is -46 kJ/mol, we can use this value to calculate the enthalpy change for the reaction.
The balanced equation shows that 2 moles of NH3 react to form 1 mole of N2 and 3 moles of H2.
Using the stoichiometric coefficients, we can calculate the enthalpy change as follows:
Enthalpy change = (1 mol N2 × enthalpy of formation of N2) + (3 mol H2 × enthalpy of formation of H2) - (2 mol NH3 × enthalpy of formation of NH3)
Substituting the given values, we get:
Enthalpy change = (1 × 0 kJ/mol) + (3 × 0 kJ/mol) - (2 × -46 kJ/mol) Enthalpy change = 0 kJ/mol + 0 kJ/mol + 92 kJ/mol Enthalpy change = 92 kJ/mol
Therefore, the enthalpy change for the reaction 2NH3 → N2 + 3H2 is 92 kJ.
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