SO3 Decomposition: Enthalpy Calculation Explained

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Hey guys! Let's dive into a chemistry problem where we need to figure out the enthalpy change (ΔH{\Delta H}) for the decomposition of sulfur trioxide (SO3). This is a classic thermochemistry problem, and we're going to break it down step by step. Essentially, we want to find out how much energy is either released or absorbed when SO3 breaks down into sulfur dioxide (SO2) and oxygen (O2).

Understanding the Reaction

The reaction we're dealing with is:

SO3(g)→SO2(g)+12O2(g){ SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) }

To solve this, we'll use Hess's Law, which states that the enthalpy change of a reaction is independent of the pathway taken. This means we can use a series of thermochemical equations to find the ΔH{\Delta H} for our reaction.

Breaking Down Hess's Law

Hess's Law is super useful because it lets us calculate enthalpy changes for reactions that are hard to measure directly. Imagine you want to find out the ΔH{\Delta H} for a complex reaction. Instead of doing the experiment in one go (which might be tricky or impossible), you can break it down into simpler steps, measure the ΔH{\Delta H} for each step, and then add them up. The total ΔH{\Delta H} will be the same as if you did the reaction in one step!

Why Enthalpy Matters

Enthalpy, represented by the symbol H, is a thermodynamic property of a system. It's essentially the sum of the internal energy of the system plus the product of its pressure and volume: H = U + PV. The enthalpy change, ΔH{\Delta H}, tells us how much heat is absorbed or released during a reaction at constant pressure.

  • If ΔH{\Delta H} is negative, the reaction is exothermic, meaning it releases heat. Think of a cozy campfire – it gives off heat, so it's exothermic. 🔥
  • If ΔH{\Delta H} is positive, the reaction is endothermic, meaning it absorbs heat. Like melting ice – it needs heat from the surroundings to melt. 🧊

Thermochemical Equations: The Building Blocks

Thermochemical equations are balanced chemical equations that also include the enthalpy change (ΔH{\Delta H}) for the reaction. These equations are crucial because they give us the energy information needed to apply Hess's Law.

For example, if we have the thermochemical equation:

A+B→CΔH=−100 kJ{ A + B \rightarrow C \quad \Delta H = -100 \text{ kJ} }

This tells us that when A and B react to form C, 100 kJ of heat is released (because ΔH{\Delta H} is negative). We can use these equations like puzzle pieces to find the ΔH{\Delta H} for our target reaction.

Using Given Thermochemical Equations

Let's assume we have the following thermochemical equations (these are examples, you'd need the actual equations from the problem):

  1. S(s)+O2(g)→SO2(g)ΔH1=−296 kJ{ S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H_1 = -296 \text{ kJ} }
  2. S(s)+32O2(g)→SO3(g)ΔH2=−395 kJ{ S(s) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) \quad \Delta H_2 = -395 \text{ kJ} }

Our goal is to manipulate these equations so that when we add them together, we get our target reaction:

SO3(g)→SO2(g)+12O2(g){ SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) }

Step-by-Step Manipulation

  1. Reverse the second equation: When we reverse an equation, we also change the sign of ΔH{\Delta H}:

    SO3(g)→S(s)+32O2(g)ΔH−2=+395 kJ{ SO_3(g) \rightarrow S(s) + \frac{3}{2} O_2(g) \quad \Delta H_{-2} = +395 \text{ kJ} }

  2. Keep the first equation as is:

    S(s)+O2(g)→SO2(g)ΔH1=−296 kJ{ S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H_1 = -296 \text{ kJ} }

  3. Add the equations: Now, add the reversed second equation and the first equation:

    SO3(g)+S(s)+O2(g)→S(s)+32O2(g)+SO2(g){ SO_3(g) + S(s) + O_2(g) \rightarrow S(s) + \frac{3}{2} O_2(g) + SO_2(g) }

    Notice that S(s){ S(s) } cancels out on both sides. Also, O2(g){ O_2(g) } on the left cancels out with 32O2(g){ \frac{3}{2} O_2(g) } on the right, leaving 12O2(g){ \frac{1}{2} O_2(g) } on the right.

    This simplifies to:

    SO3(g)→SO2(g)+12O2(g){ SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) }

    Which is exactly the reaction we wanted!

  4. Add the enthalpy changes: Add the ΔH{\Delta H} values for the manipulated equations:

    ΔH=ΔH−2+ΔH1=+395 kJ+(−296 kJ)=99 kJ{ \Delta H = \Delta H_{-2} + \Delta H_1 = +395 \text{ kJ} + (-296 \text{ kJ}) = 99 \text{ kJ} }

Therefore, the enthalpy change for the decomposition of sulfur trioxide is 99 kJ.

Important Considerations

State Symbols Matter

Always pay attention to the state symbols (e.g., (s) for solid, (l) for liquid, (g) for gas, (aq) for aqueous). The enthalpy change can be different for different states. For example, the ΔH{\Delta H} for vaporizing water (H2O(l) → H2O(g)) is significant and must be considered in any calculation involving water in different states.

Standard Conditions

Enthalpy changes are often given under standard conditions (298 K and 1 atm pressure). These are denoted with a superscript ° (e.g., ΔH°{\Delta H°}). Make sure you're using the correct values for the conditions specified in the problem.

Common Mistakes to Avoid

  • Forgetting to reverse the sign of ΔH{\Delta H} when reversing an equation. This is a very common mistake, so double-check your signs!
  • Not multiplying ΔH{ \Delta H } by the coefficient when multiplying an equation. If you multiply an entire equation by a factor (e.g., to balance it), you must also multiply the ΔH{ \Delta H } by the same factor.
  • Ignoring state symbols. As mentioned earlier, state symbols matter. Make sure you're using the correct enthalpy values for the correct states.
  • Mixing up reactants and products. Always double-check that you're adding the equations in the correct direction to get your target reaction.

Example

Let's say we are given the following:

  1. 2SO2(g)+O2(g)→2SO3(g){2SO_2(g) + O_2(g) \rightarrow 2SO_3(g) } ΔH=−198 kJ{\Delta H = -198 \text{ kJ}}

We want to find the ΔH{\Delta H} for:

SO3(g)→SO2(g)+12O2(g){SO_3(g) \rightarrow SO_2(g) + \frac{1}{2}O_2(g)}

We need to reverse the first equation and divide it by 2:

SO3(g)→SO2(g)+12O2(g){SO_3(g) \rightarrow SO_2(g) + \frac{1}{2}O_2(g)} ΔH=+99 kJ{\Delta H = +99 \text{ kJ}}

Conclusion

Finding the enthalpy change for the decomposition of sulfur trioxide involves using Hess's Law and manipulating thermochemical equations. By reversing equations, adding them together, and adjusting the ΔH{\Delta H} values accordingly, we can determine the overall enthalpy change for the reaction. Remember to pay attention to state symbols and standard conditions to ensure accurate calculations. Keep practicing, and you'll master these thermochemistry problems in no time! You got this!