Question

Calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ given the following data: \(\begin{array}{ll}\text { Equation } & \Delta H(\mathrm{~kJ}) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) \rightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & -1010 \\\ \mathrm{~N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -143 \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\end{array}\)

Short answer

The \(\Delta H\) for the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) is approximately \(-2041.4~\mathrm{kJ}\).

Step-by-step solution

Identify relevant reactions

Check which given reactions contain the species involved in the desired reaction. All four of the given reactions have at least one species in common with the desired reaction. Step 2: Eliminate unwanted species

Eliminate unwanted species

We want to remove unwanted species present in the given reactions by adding or subtracting reactions to cancel out these species. We can eliminate \(\mathrm{NH}_3\) and \(\mathrm{N}_2\mathrm{O}\) by combining the first three given reactions. Add the first and second given reactions: \(2 \mathrm{NH}_3(g) + 3 \mathrm{N}_2\mathrm{O}(g) \rightarrow 4 \mathrm{N}_2(g) + 3 \mathrm{H}_2\mathrm{O}(l) \quad (\Delta H_1 = -1010~\mathrm{kJ})\) \(\mathrm{N}_2\mathrm{O}(g) + 3 \mathrm{H}_2(g) \rightarrow \mathrm{N}_2\mathrm{H}_4(l) + \mathrm{H}_2\mathrm{O}(l) \quad (\Delta H_2 = -317~\mathrm{kJ})\) Resulting reaction: \(2 \mathrm{NH}_3(g) + 4 \mathrm{N}_2\mathrm{O}(g) + 3 \mathrm{H}_2(g) \rightarrow 4 \mathrm{N}_2(g) + \mathrm{N}_2\mathrm{H}_4(l) + 4 \mathrm{H}_2\mathrm{O}(l)\) The sum of the first and second reaction's \(\Delta H\) values: \(\Delta H_{1+2} = -1327~\mathrm{kJ}\) Then subtract the third reaction from the result: \(2 \mathrm{NH}_3(g) + 4 \mathrm{N}_2\mathrm{O}(g) + 3 \mathrm{H}_2(g) \rightarrow 4 \mathrm{N}_2(g) + \mathrm{N}_2\mathrm{H}_4(l) + 4 \mathrm{H}_2\mathrm{O}(l) \quad (\Delta H_{1+2} = -1327~\mathrm{kJ})\) \(-[2 \mathrm{NH}_3(g) + 0.5 \mathrm{O}_2(g) \rightarrow \mathrm{N}_2\mathrm{H}_4(l) + \mathrm{H}_2\mathrm{O}(l)] \quad (\Delta H_3 = 143~\mathrm{kJ})\) (Note that the sign of \(\Delta H\) is inverted here, since we are subtracting the reaction.) Resulting reaction: \(-0.5 \mathrm{O}_2(g) + 4 \mathrm{N}_2\mathrm{O}(g) + 3 \mathrm{H}_2(g) \rightarrow 4 \mathrm{N}_2(g) + 3 \mathrm{H}_2\mathrm{O}(l)\) The final \(\Delta H\) is the difference of the previous reactions' \(\Delta H\) values: \(\Delta H_{result} = -1327 + 143 = -1184~\mathrm{kJ}\) Step 3: Obtaining the desired reaction

Obtaining the desired reaction

Add the fourth given reaction twice to the resulting reaction obtained in step 2, to obtain the desired reaction. \(3 \mathrm{H}_2(g) + \mathrm{O}_2(g) \rightarrow 3 \mathrm{H}_2\mathrm{O}(l) \quad (\Delta H_4 = -285.8 * 3~\mathrm{kJ})\) (The fourth given reaction is multiplied by 3 since we require 3 moles of \(\mathrm{H}_2\mathrm{O}\) in the final reaction.) Add this reaction to the resulting reaction in step 2: \(-0.5 \mathrm{O}_2(g) + 4 \mathrm{N}_2\mathrm{O}(g) + 3 \mathrm{H}_2(g) \rightarrow 4 \mathrm{N}_2(g) + 3 \mathrm{H}_2\mathrm{O}(l) \quad (\Delta H_{result} = -1184~\mathrm{kJ})\) \(+~3 \mathrm{H}_2(g) + \mathrm{O}_2(g) \rightarrow 3 \mathrm{H}_2\mathrm{O}(l) \quad (\Delta H_4 = -285.8 * 3~\mathrm{kJ})\) Resulting reaction: \(\mathrm{N}_2\mathrm{H}_4(l) + \mathrm{O}_2(g) \rightarrow \mathrm{N}_2(g) + 2 \mathrm{H}_2\mathrm{O}(l)\) The final \(\Delta H\) is the sum of previous reactions' \(\Delta H\) values: \(\Delta H_{final} = -1184 + (-285.8 * 3) = -2041.4~\mathrm{kJ}\) The \(\Delta H\) for the desired reaction is approximately \(-2041.4~\mathrm{kJ}\).

Key concepts

Hess's Law

Hess's Law is a fundamental principle in thermochemistry that allows us to calculate the enthalpy change (ΔH) of a chemical reaction even when it can’t be measured directly. Instead of measuring ΔH for the reaction directly, we use a series of steps, adding the known enthalpy changes from related reactions. The beauty of Hess's Law is that we can manipulate these individual steps, such as reversing them or scaling them, and then sum all of their enthalpy changes.

• This law is based on the principle that enthalpy is a state function: It depends only on the initial and final states, not the path taken.
• By creating a path using known reactions, one can computationally determine the enthalpy change for complex reactions.

In the given exercise, we use Hess's Law to find the ΔH for the reaction of hydrazine ( N_<2H_4 d4>) with oxygen by combining other reactions with known enthalpy changes. This involves steps such as adding or subtracting reactions to align with the target reaction, ensuring the conservation of mass and energy.

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the heat changes that occur during chemical reactions. Understanding the enthalpy change, denoted as ΔH, is crucial, as it tells us whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).

• An exothermic reaction has a negative ΔH because energy is released into the surroundings, as seen in most combustion reactions.
• An endothermic reaction has a positive ΔH because it absorbs energy from the surroundings, like in photosynthesis.

In our specific exercise, we calculate the ΔH for a reaction involving hydrazine and oxygen, revealing that it is highly exothermic with a ΔH of approximately -2041.4 kJ. This suggests that a substantial amount of energy is released during the transformation to nitrogen gas and water.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products through breaking existing bonds and forming new ones. For any reaction, it’s important to consider both stoichiometry and thermodynamics.

• Stoichiometry ensures that equations are balanced, indicating equal numbers of atoms for each element on both sides of the equation.
• Thermodynamic principles, like enthalpy calculations, help predict whether a reaction will occur spontaneously under given conditions.

In the exercise, we encounter a multi-step process to find the overall energy change for a reaction where hydrazine reacts with oxygen to form nitrogen gas and water. Understanding the reactants, intermediates, and products ensures that the reaction path aligns with our desired reaction, enabling us to correctly apply Hess's Law and reach our goal efficiently.