Problem 120 - Chemistry An Atoms First Approach

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: $$\begin{array}{lr}\text { Equation } & \Delta H(\mathrm{kJ}) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & -1010 \\ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -143 \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & -286 \end{array}$$

Step-by-Step Solution

Answer

The enthalpy change for the given reaction, \(N_2 H_4(l) + O_2(g) \rightarrow N_2(g) + 2 H_2 O(l)\), is 143 kJ.

Step 1: Find the correct equation for N2H4(l)

In the given data, the equation involving N2H4(l) is: \( 2NH_3(g) + \frac{1}{2} O_2(g) \rightarrow N_2 H_4(l) + H_2 O(l) \) The ΔH value for this equation is -143 kJ.

Step 2: Find the correct equation for O2(g)

We need a reaction where O2(g) stays the same and only one instance is required. The following equation satisfies this condition: \(H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2 O(l)\) The ΔH value for this equation is -286 kJ.

Step 3: Manipulate the equations to get N2(g) + 2 H2O(l)

The correct equation to obtain N2(g) can be found by combining both obtained equations: \( N_2 H_4(l) + H_2 O(l) + H_2(g) + \frac{1}{2} O_2(g) \rightarrow N_2 H_4(l) + 2 H_2 O(l) \)

Step 4: Calculate the ΔH value for the desired equation

To get the desired equation, subtract the second equation from the first one: \(N_2 H_4(l) + O_2(g) \rightarrow N_2(g) + 2 H_2 O(l)\) Now, we need to find the ΔH value for this new equation. To find the enthalpy change for the desired reaction, subtract the ΔH value of the second equation (-286 kJ) from the ΔH value of the first equation (-143 kJ): ΔH (desired equation) = ΔH (first equation) - ΔH (second equation) ΔH = -143 kJ - (-286 kJ) ΔH = 143 kJ So, the enthalpy change for the given reaction is 143 kJ.

Key Concepts

Chemical Reaction Energetics

Understanding the energetics of chemical reactions is pivotal for students of chemistry. Chemical reaction energetics involves the study of energy changes that occur during reactions. The key term in this concept is 'enthalpy' (\( \text{H} \)), which is a measure of the total energy of a thermodynamic system.

When a reaction occurs, energy is either absorbed or released, and we refer to this as the enthalpy change (\( \text{\text{ΔH}} \)), which can be either positive or negative. A negative \( \text{\text{ΔH}} \) indicates that the reaction is exothermic, meaning it releases heat into the surroundings. Conversely, a positive \( \text{\text{ΔH}} \) suggests that the reaction is endothermic, absorbing heat from the surroundings.

To illustrate, in our reaction: \[ \text{N}_2\text{H}_4(l) + \text{O}_2(g) \rightarrow \text{N}_2(g) + 2 \text{H}_2\text{O}(l) \] it is the change in enthalpy that we are interested in calculating to understand the energy dynamics involved. This understanding is critical for predicting reaction behavior, as well as for the practical design of chemical processes.

Hess's Law

When it comes to calculating enthalpy changes, Hess's law is a valuable principle. It states that the total enthalpy change for a chemical reaction is the same, no matter how the reaction occurs in a series of steps. This principle is based on the law of conservation of energy and allows us to calculate the enthalpy change of a reaction by using known enthalpy changes of related reactions.

In the context of our problem, we do not have a direct way to measure the enthalpy change for the given reaction, but we can use Hess's Law to piece together information from related reactions to find the desired value. By manipulating and combining known equations, and ensuring that they add up to the desired equation, we can sum the enthalpy changes of these steps to find the overall enthalpy change.

Applying Hess's Law

By analyzing the given chemical equations, we looked for those that, when combined appropriately, would yield the target equation. We aligned the molecules and properly adjusted their stoichiometry, ensuring the coefficients of reactants and products would match the desired reaction. Subtracting the enthalpies of these equations as done in the provided solution reflects the direct application of Hess's Law.

Standard Enthalpy of Formation

To further our understanding of chemical energetics, we delve into the concept of standard enthalpy of formation (\( \text{ΔH}_f^\text{o} \)). This is defined as the change in enthalpy that accompanies the formation of 1 mole of a substance from its constituent elements under standard conditions (1 atm pressure and 298.15 K temperature).

Standard enthalpies of formation are fundamental in calculating the enthalpy changes for reactions, especially when direct measurement is not possible. Each pure element in its standard state (like O2(g) or N2(g) at 25°C and 1 atmosphere) is assigned a standard enthalpy of formation of zero since there is no change involved in 'forming' an element from itself.

Importance in Calculations

In computations, we often use a table of standard enthalpies of formation for various compounds. These values are carefully measured under standard conditions. When we want to determine the enthalpy change for a reaction, we can use the known standard enthalpies of formation for the reactants and products. Such values offer a baseline for comparing the energetics of different chemical reactions and designing new synthetic pathways in chemistry.