Thermochemical Equations: Formation, Combustion, Decomposition

Hey guys! Let's dive into the fascinating world of thermochemistry and break down how to write thermochemical equations. We’ll tackle formation, combustion, and decomposition reactions step by step. So, grab your calculators and let's get started!

Understanding Thermochemical Equations

First off, what exactly is a thermochemical equation? Simply put, it's a chemical equation that includes the enthalpy change ($\Delta H$) for the reaction. This value tells us whether the reaction releases heat (exothermic, $\Delta H$ is negative) or absorbs heat (endothermic, $\Delta H$ is positive). Writing these equations correctly is crucial for understanding energy changes in chemical reactions. We need to make sure we balance the equations, indicate the physical states of reactants and products (solid (s), liquid (l), gas (g), or aqueous (aq)), and include the correct $\Delta H$ value with its sign and units (usually kJ/mol).

Key Components of a Thermochemical Equation

To get this right, there are a few key things we need to include:

1. Balanced Chemical Equation: The number of atoms for each element must be the same on both sides of the equation. This ensures we’re following the law of conservation of mass.
2. Physical States: Indicate whether each substance is a solid (s), liquid (l), gas (g), or aqueous solution (aq). The enthalpy change can vary significantly depending on the state.
3. Enthalpy Change ($\Delta H$): This is the amount of heat absorbed or released during the reaction at constant pressure. It's typically given in kJ/mol, indicating the heat change per mole of reaction as it's written.
4. Standard Conditions: For standard enthalpy changes ($\Delta H^\circ$), reactions are assumed to occur at standard conditions (298 K and 1 atm). This is important for comparing enthalpy changes between different reactions.

By including these components, we create a clear and comprehensive picture of the energy changes occurring in a chemical reaction. Now, let's dive into some specific examples.

a. Heat of Formation of $NH_4Cl(s)$ (-314.4 kJ/mol)

The standard heat of formation, denoted as $\Delta H_f^\circ$, is the enthalpy change when one mole of a compound is formed from its elements in their standard states. Standard states are the most stable form of the element at 298 K (25°C) and 1 atm pressure. For example, the standard state of nitrogen is $N_2(g)$, hydrogen is $H_2(g)$, and chlorine is $Cl_2(g)$.

So, for the formation of ammonium chloride ($NH_4Cl(s)$), we need to start with its constituent elements in their standard states: nitrogen ($N_2$), hydrogen ($H_2$), and chlorine ($Cl_2$). The balanced equation will show these elements reacting to form one mole of $NH_4Cl(s)$. Since the heat of formation is given as -314.4 kJ/mol, this means the reaction is exothermic – it releases heat.

Step-by-Step Process

1. Identify the elements: We have nitrogen (N), hydrogen (H), and chlorine (Cl).
2. Write the elements in their standard states: $N_2(g)$, $H_2(g)$, and $Cl_2(g)$.
3. Write the product: $NH_4Cl(s)$.
4. Write the unbalanced equation: $N_2(g) + H_2(g) + Cl_2(g) \rightarrow NH_4Cl(s)$
5. Balance the equation: We need 1 nitrogen, 4 hydrogens, and 1 chlorine on both sides. $N_2(g) + 4H_2(g) + Cl_2(g) \rightarrow 2NH_4Cl(s)$ Since we want the equation for one mole of $NH_4Cl$, we divide the entire equation by 2: $\frac{1}{2}N_2(g) + 2H_2(g) + \frac{1}{2}Cl_2(g) \rightarrow NH_4Cl(s)$
6. Include the enthalpy change: Since the standard heat of formation is -314.4 kJ/mol, we add this to our equation.

The Thermochemical Equation

The final thermochemical equation for the formation of $NH_4Cl(s)$ is:

$\frac{1}{2}N_2(g) + 2H_2(g) + \frac{1}{2}Cl_2(g) \rightarrow NH_4Cl(s) \quad \Delta H_f^\circ = -314.4 \text{ kJ/mol}$

This equation tells us that when one mole of $NH_4Cl(s)$ is formed from its elements in their standard states, 314.4 kJ of heat is released.

b. Standard Enthalpy of Combustion of Methane (-820 kJ/mol)

Next up is combustion. The standard enthalpy of combustion, $\Delta H_c^\circ$, is the enthalpy change when one mole of a substance completely reacts with oxygen under standard conditions. For hydrocarbons like methane ($CH_4$), combustion produces carbon dioxide ($CO_2$) and water ($H_2O$).

The standard enthalpy of combustion for methane is given as -820 kJ/mol, indicating that this is an exothermic reaction – burning methane releases a lot of heat, which is why it's used as a fuel! We need to write the balanced equation for the reaction of one mole of methane with oxygen to produce carbon dioxide and water, and then include the enthalpy change.

Step-by-Step Process

1. Identify the reactants: Methane ($CH_4$) and oxygen ($O_2$).
2. Identify the products: Carbon dioxide ($CO_2$) and water ($H_2O$).
3. Write the unbalanced equation: $CH_4(g) + O_2(g) \rightarrow CO_2(g) + H_2O(l)$
4. Balance the equation:
   • Start by balancing the carbons: 1 carbon on each side, so that's good.
   • Next, balance the hydrogens: 4 hydrogens on the left, so we need 2 water molecules on the right. $CH_4(g) + O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$
   • Finally, balance the oxygens: 2 oxygens in $CO_2$ and 2 oxygens in $2H_2O$, so we have 4 oxygens on the right. We need 2 oxygen molecules on the left. $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$
5. Include the enthalpy change: The standard enthalpy of combustion is -820 kJ/mol.

The Thermochemical Equation

The final thermochemical equation for the combustion of methane is:

$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H_c^\circ = -820 \text{ kJ/mol}$

This equation shows that when one mole of methane is completely burned in oxygen, 820 kJ of heat is released.

c. Decomposition of 2 mol of $H_2O(l)$ ($\Delta H^\circ H_2O(l)$ = -286 kJ/mol)

Now, let's tackle decomposition. Decomposition is essentially the reverse of formation. It's the process where a compound breaks down into its elements. In this case, we're looking at the decomposition of water ($H_2O$) into hydrogen ($H_2$) and oxygen ($O_2$).

The standard enthalpy of formation for water is given as -286 kJ/mol. Remember, decomposition is the reverse process, so the enthalpy change for decomposition will have the same magnitude but the opposite sign. Since we’re decomposing 2 moles of water, we need to consider that in our calculation.

Step-by-Step Process

1. Identify the reactant: Water ($H_2O(l)$).

2. Identify the products: Hydrogen ($H_2(g)$) and oxygen ($O_2(g)$).

3. Write the unbalanced equation: $H_2O(l) \rightarrow H_2(g) + O_2(g)$

4. Balance the equation:

   • We have 2 hydrogens on both sides, so that’s balanced.
   • We have 1 oxygen on the left and 2 on the right, so we need to balance the oxygens. Put a coefficient of $\frac{1}{2}$ in front of $O_2$. $H_2O(l) \rightarrow H_2(g) + \frac{1}{2}O_2(g)$

5. Consider the amount: We are decomposing 2 moles of $H_2O$, so multiply the entire equation by 2: $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$

6. Calculate the enthalpy change:

   • The enthalpy of formation of $H_2O(l)$ is -286 kJ/mol.
   • The enthalpy of decomposition of $H_2O(l)$ (for 1 mole) is +286 kJ/mol (opposite sign).
   • Since we're decomposing 2 moles, multiply by 2: $\Delta H = 2 \times +286 \text{ kJ/mol} = +572 \text{ kJ}$

The Thermochemical Equation

The final thermochemical equation for the decomposition of 2 moles of $H_2O(l)$ is:

$2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \quad \Delta H = +572 \text{ kJ}$

This equation tells us that when 2 moles of liquid water are decomposed into hydrogen and oxygen, 572 kJ of heat is absorbed, making it an endothermic process.

Mastering Thermochemical Equations

Alright, guys, we've covered quite a bit! Writing thermochemical equations might seem daunting at first, but breaking it down step by step makes it much easier. Remember to:

• Balance the equation: Ensure the number of atoms for each element is the same on both sides.
• Include physical states: Specify whether substances are solid (s), liquid (l), gas (g), or aqueous (aq).
• Add the enthalpy change: Include the $\Delta H$ value with the correct sign and units.
• Consider standard conditions: Use standard states for elements when dealing with standard enthalpy changes.

By following these steps, you’ll be writing thermochemical equations like a pro in no time! Keep practicing, and you'll get the hang of it. Chemistry can be super fun once you understand the basics!