Question

Calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ \begin{aligned} 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) & \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1010 . \mathrm{kJ} \\ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-317 \mathrm{~kJ} \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-143 \mathrm{~kJ} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \Delta H &=-286 \mathrm{~kJ} \end{aligned} $$

Short answer

The enthalpy change for the given reaction, N₂H₄(l) + O₂(g) → N₂(g) + 2 H₂O(l), is 836 kJ.

Step-by-step solution

Identify reactions that can be combined to form the desired reaction

The reactions given are: 1. 2 NH₃(g) + 3 N₂O(g) → 4 N₂(g) + 3 H₂O(l), \(\Delta H = -1010 \: kJ\) 2. N₂O(g) + 3 H₂(g) → N₂H₄(l) + H₂O(l), \(\Delta H = -317 \: kJ\) 3. 2 NH₃(g) + ½ O₂(g) → N₂H₄(l) + H₂O(l) , \(\Delta H = -143 \: kJ\) 4. H₂(g) + ½ O₂(g) → H₂O(l), \(\Delta H = -286 \: kJ\) By observing these reactions, we can see that reaction 2 needs to be added to the reverse of reaction 1, and then reaction 3 needs to be subtracted.

Reverse reaction 1 and compute its new enthalpy change

Reverse reaction 1: 4 N₂(g) + 3 H₂O(l) → 2 NH₃(g) + 3 N₂O(g) Now, compute the new enthalpy change: \(\Delta H' = -(-1010 \: kJ) = 1010 \: kJ\) The new reaction and its enthalpy change are: 4 N₂(g) + 3 H₂O(l) → 2 NH₃(g) + 3 N₂O(g), \(\Delta H' = 1010 \: kJ\)

Add reverse of reaction 1 and reaction 2

Add the new reaction 1 and reaction 2: (4 N₂(g) + 3 H₂O(l) → 2 NH₃(g) + 3 N₂O(g)) + (N₂O(g) + 3 H₂(g) → N₂H₄(l) + H₂O(l)) This gives us the new reaction: 4 N₂(g) + 4 H₂O(l) + 3 H₂(g) → 2 NH₃(g) + N₂H₄(l) + H₂O(l) Compute the enthalpy change: \(\Delta H'' = \Delta H' + \Delta H_2 = 1010 \: kJ - 317 \: kJ = 693 \: kJ\)

Subtract reaction 3 from the result obtained in step 3

Subtract reaction 3 from the result in step 3: (4 N₂(g) + 4 H₂O(l) + 3 H₂(g) → 2 NH₃(g) + N₂H₄(l) + H₂O(l)) - (2 NH₃(g) + ½ O₂(g) → N₂H₄(l) + H₂O(l)) This gives us the desired reaction: 3 H₂(g) + ½ O₂(g) → N₂(g) + 2 H₂O(l) Compute the enthalpy change: \(\Delta H_\mathrm{desired} = \Delta H'' - \Delta H_3 = 693 \: kJ - (-143 \: kJ) = 836 \: kJ\)

Conclusion

The enthalpy change for the given reaction, N₂H₄(l) + O₂(g) → N₂(g) + 2 H₂O(l), is 836 kJ.

Key concepts

Chemical Reaction Enthalpy

In thermochemistry, the enthalpy change, denoted by \( \Delta H \), is a measure of heat absorbed or released during a chemical reaction under constant pressure. It is defined as the difference between the enthalpy of the products and the reactants. The sign of \( \Delta H \) indicates whether a reaction is exothermic (releases heat, \( \Delta H < 0 \) ) or endothermic (absorbs heat, \( \Delta H > 0 \) ). Enthalpy change calculations are fundamental in predicting the energy changes involved in chemical processes.

Understanding enthalpy change aids in the study of energy flows in reactions, which is essential for applications like designing energy-efficient synthesis pathways and assessing the feasibility of industrial processes. Consider a balanced chemical equation; the coefficients indicate the stoichiometric quantities, which are used to calculate the total enthalpy change for the reaction.

Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps the reaction is carried out in. This principle is crucial because it allows us to calculate the enthalpy change of a reaction even if it cannot be measured directly. Instead, we can use known enthalpy changes of related reactions to find the desired value.

To apply Hess's Law, one may need to manipulate the given reactions by reversing, multiplying, or dividing them by a coefficient to sum up to the target reaction. The enthalpy changes for these manipulated reactions must also be adjusted accordingly (reversed reactions have their \( \Delta H \) values reversed in sign, and multiplying or dividing a reaction by a number will also multiply or divide its \( \Delta H \) by the same number). The key is to ensure that all extraneous substances cancel out, leaving only the desired reaction equation.

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the energy and heat associated with chemical reactions and physical transformations. It encompasses the principles of thermodynamics to understand how energy is transformed and transferred in chemical processes, and to assess the spontaneity and feasibility of reactions.

The First Law of Thermodynamics, which states that energy cannot be created or destroyed, only converted from one form to another, is the bedrock of thermochemistry. Enthalpy is one of the central concepts in this field and is especially important when considering reactions at constant pressure, which is common in most laboratory and industrial conditions. Thermochemistry also deals with the concepts of entropy, Gibbs free energy, and the heat capacities of substances, all of which give insight into different aspects of a system's energy and spontaneous behavior.