calculate the enthalpy change for the reaction n2h4(l) + 02(g) --> n2(g) + 2h20(l)
To calculate the enthalpy change for a reaction, you need to use the enthalpy values of the reactants and products involved. The enthalpy change is determined by the difference between the total enthalpy of the products and the total enthalpy of the reactants.
To find the enthalpy change for the given reaction:
1. Look up the standard enthalpy of formation (ΔHf) for each compound involved in the reaction. The standard enthalpy of formation represents the enthalpy change when one mole of a compound is formed from its elements in their standard states at a specific temperature and pressure.
The standard enthalpy of formation values for the compounds involved in the reaction are:
ΔHf(N2H4) = +95.0 kJ/mol
ΔHf(O2) = 0 kJ/mol
ΔHf(N2) = 0 kJ/mol
ΔHf(H2O) = -285.8 kJ/mol (Note: The value is per mole of H2O, and since we have 2 moles of H2O in the products, we multiply the value by 2)
2. Determine the stoichiometric coefficients in the balanced chemical equation. In this case, the coefficients are:
N2H4: 1
O2: 1
N2: 1
H2O: 2
3. Calculate the enthalpy change by using the following equation:
ΔH = Σ(n * ΔHf(products)) - Σ(n * ΔHf(reactants))
Σ indicates the sum, n represents the stoichiometric coefficient, ΔHf is the standard enthalpy of formation.
Substituting the values into the equation, we get:
ΔH = (1 * ΔHf(N2) + 2 * ΔHf(H2O)) - (1 * ΔHf(N2H4) + 1 * ΔHf(O2))
Plugging in the values:
ΔH = (0 kJ/mol + 2 * (-285.8 kJ/mol)) - (1 * 95.0 kJ/mol + 0 kJ/mol)
ΔH = (-571.6 kJ/mol) - (95.0 kJ/mol)
ΔH = -666.6 kJ/mol
Therefore, the enthalpy change (ΔH) for the reaction is -666.6 kJ/mol.
Using Gibbs Relationship...
2*heatformation H2O(l)-HeatformationN2H4=enthalpy