Question

Nitrogen gas and hydrogen gas react to form ammonia according to the following thermochemical equation. 3H2+N2⟶2NH3ΔH=−96kJ Write the balanced chemical equation for a reaction involving these substances with the following change in enthalpy. ΔH=192kJ Enter a balanced chemical equation. Do NOT enter phase designations.

Answer 1

6H2+2N2⟶4NH3ΔH=−192kJ

Step-by-step explanation:

192/96=2 so we will obtain twice the energy then we need twice the equivalents and twice the coefficients.

Answer 2

4 NH₃ ⟶ 6 H₂ + 2 N₂ ΔH= 192 kJ

Step-by-step explanation:

Let's consider the following thermochemical equation.

3 H₂ + N₂ ⟶ 2 NH₃ ΔH= −96 kJ

According to Lavoisier-Laplace law, if we reverse this reaction, its enthalpy will have the same value and opposite sign than that of the direct reaction.

2 NH₃ ⟶ 3 H₂ + N₂ ΔH= 96 kJ

Enthalpy is an extensive property, that is, it depends on the amount of matter. If we double the amount of matter (by multiplying the stoichiometric coefficients by 2), the enthalpy will also be doubled.

4 NH₃ ⟶ 6 H₂ + 2 N₂ ΔH= 192 kJ