Write the balanced chemical reaction (showing appropriate symbols and states) for the chemical reaction with enthalpy change equal to ΔHf° [NH3(g)]
Calculate the enthalpy for this reaction:
2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ
Given the following thermochemical equations:
C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1299.5 kJ
C(s) + O2(g) ---> CO2(g) ΔH° = -393.5 kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ
2CO2(g) + H2O(ℓ) ---> C2H2(g) + (5/2)O2(g) ΔH° = +1299.5 kJ
2C(s) + 2O2(g) ---> 2CO2(g) ΔH° = -787 kJ
H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ
+1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ Good Luck
To determine the balanced chemical reaction with an enthalpy change equal to ΔHf° (standard enthalpy of formation) for ammonia (NH3) in the gaseous state, we need to consider the reactants and products involved.
The standard enthalpy of formation (ΔHf°) is the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states, with all reactants and products also being in their standard states.
For ammonia (NH3), the reactants are nitrogen gas (N2) in the gaseous state and hydrogen gas (H2) in the gaseous state. The product is ammonia (NH3) in the gaseous state.
The balanced chemical reaction is as follows:
N2(g) + 3H2(g) → 2NH3(g)
This reaction shows that one molecule of nitrogen gas (N2) reacts with three molecules of hydrogen gas (H2) to produce two molecules of ammonia (NH3) in the gaseous state. The coefficients in front of each compound represent the stoichiometric ratio required for a balanced reaction.
Note: The ΔHf° value for NH3 is usually given as -46.1 kJ/mol, indicating the enthalpy change when one mole of NH3 is formed.