Problem 74 - General Chemistry

Question

Hydrogen cyanide is a highly poisonous, volatile liquid. It can be prepared by the reaction $$ \mathrm{CH}_{4}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{HCN}(g)+3 \mathrm{H}_{2}(g) $$ What is the heat of reaction at constant pressure? Use the following thermochemical equations: $$ \begin{gathered} \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) ; \Delta H=-91.8 \mathrm{~kJ} \\ \mathrm{C}(\text { graphite })+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{4}(g) ; \Delta H=-74.9 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+2 \mathrm{C}(\text { graphite })+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{HCN}(g) \\ \Delta H=270.3 \mathrm{~kJ} \end{gathered} $$

Step-by-Step Solution

Answer

The heat of reaction is \(-14.35 \, \mathrm{kJ} \) at constant pressure.

Step 1: Write Known Reactions

List the given thermochemical equations:1. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\), \(\Delta H=-91.8 \, \mathrm{kJ} \)2. \(\mathrm{C}(\text{graphite})+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{4}(g)\), \(\Delta H=-74.9 \, \mathrm{kJ} \)3. \(\mathrm{H}_{2}(g)+2 \mathrm{C}(\text{graphite})+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{HCN}(g)\), \(\Delta H=270.3 \, \mathrm{kJ} \).

Step 2: Reverse Reactions for Target Equation

We need \( \mathrm{HCN}(g) \) on the reactant side in our target equation, so reverse the third equation:\(2 \mathrm{HCN}(g) \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{C}(\text{graphite})+\mathrm{N}_{2}(g)\) with a change of sign, \( \Delta H = -270.3 \, \mathrm{kJ} \).

Step 3: Adjust Reaction Stoichiometry

Divide the reversed third reaction by 2 to match \( \mathrm{HCN} \) in the equation:\( \mathrm{HCN}(g) \longrightarrow 0.5 \mathrm{H}_{2}(g)+\mathrm{C}(\text{graphite})+0.5 \mathrm{N}_{2}(g)\), \( \Delta H = -135.15 \, \mathrm{kJ} \).

Step 4: Balance Hydrogen in Reactions

Adjust the first reaction for 1 ammonia:\(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)\), \( \Delta H = -45.9 \, \mathrm{kJ} \).

Step 5: Combine Adjusted Reactions

Sum the adjusted thermochemical equations to obtain the target reaction:1. \( \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}(\text{graphite})+2 \mathrm{H}_{2}(g) \), \( \Delta H = 74.9 \, \mathrm{kJ} \), reversed2. \( \mathrm{NH}_{3}(g) \longrightarrow \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \), \( \Delta H = 45.9 \, \mathrm{kJ} \), reversed3. \(\mathrm{HCN}(g) \longrightarrow 0.5\mathrm{H}_{2}(g)+\mathrm{C}(\text{graphite})+0.5\mathrm{N}_{2}(g) \), \( \Delta H = -135.15 \, \mathrm{kJ} \), adjustedThis combines to form \( \mathrm{CH}_{4}(g) + \mathrm{NH}_{3}(g) \longrightarrow \mathrm{HCN}(g) + 3 \mathrm{H}_{2}(g)\) with \( \Delta H = 74.9 + 45.9 - 135.15 = -14.35 \, \mathrm{kJ} \).

Step 6: Calculate the Heat of Reaction

Sum the energies of the combined reactions:\( \Delta H = 74.9 + 45.9 - 135.15 = -14.35 \, \mathrm{kJ} \).

Key Concepts

Heat of Reaction

In thermochemistry, the heat of reaction is a crucial concept. It refers to the amount of heat either absorbed or released during a chemical reaction. Think of it as the energy change that occurs when reactants transform into products. Whether heat is absorbed or released, we measure the heat of reaction at a constant pressure for consistency.

Here's a simple way to phrase it:- **Exothermic Reactions**: These release heat, resulting in a negative enthalpy change ( \( \Delta H < 0 \) ).- **Endothermic Reactions**: These absorb heat, leading to a positive enthalpy change ( \( \Delta H > 0 \)).

In the given problem, we find that the heat of reaction for forming hydrogen cyanide from methane and ammonia is \( -14.35 \, \mathrm{kJ} \). This means the reaction is slightly exothermic as it releases a small amount of heat.

Understanding the heat of reaction helps predict how much heat will be exchanged with the surroundings, indicating whether the process will feel warm or cold.

Thermochemical Equations

Thermochemical equations are chemical equations that include the associated enthalpy change ( \( \Delta H \)). They are a vital tool for understanding energy changes in reactions. These equations not only show the reactants and products but also how much heat is involved.

Key features of thermochemical equations include:- **Balanced**: They must be balanced for both mass and energy.- **State Symbols**: Indicate the physical states of substances (solid, liquid, gas, aqueous).- **Enthalpy Change**: Include the \( \Delta H \) value, often in kilojoules ( \( \mathrm{kJ} \)), with the sign indicating whether the reaction is exothermic or endothermic.

In the exercise, we work with several thermochemical equations to calculate the heat of reaction for the formation of hydrogen cyanide. Understanding these equations helps in calculating how individual reactions contribute to the overall energy change.

Hess's Law

Hess's Law is a powerful concept in thermochemistry. It states that the enthalpy change of a reaction is the same regardless of whether the reaction occurs in one step or multiple steps. This law allows us to add or subtract known enthalpy changes of multiple reactions to find the enthalpy change of a desired reaction.

Key ideas include:- **State Independence**: The total enthalpy change is independent of the path between the initial and final states.- **Algebraic Summation**: We can rearrange, multiply, or reverse known reactions and then sum their enthalpy changes to derive the target reaction's \( \Delta H \).

In our exercise, we've seen Hess's Law applied by rearranging the given thermochemical equations, reversing some reactions, and finding the heat of reaction for hydrogen cyanide formation. By meticulously adjusting and summing the known enthalpy changes, we reached the solution of \( -14.35 \, \mathrm{kJ} \) for the target equation.

Hess's Law is essential for solving complex problems where direct measurement of enthalpy change isn't feasible.