Calculating Equilibrium Constant (K) For A Reverse Reaction

Jan 12, 2026 by Tim Redaksi 60 views

Hey guys! Let's dive into a cool chemistry problem. We're given an equilibrium reaction and its equilibrium constant (K) at a specific temperature, and then we're asked to figure out the K for a reversed and modified version of that reaction. Sounds a bit tricky, right? Don't worry, it's actually pretty straightforward once you get the hang of it. We'll break it down step-by-step, making sure we cover all the important concepts. So, grab your notebooks and let's get started! Understanding this concept is crucial for anyone studying chemical kinetics and equilibrium. This is the foundation for further advanced studies. We will cover the core concepts needed to understand and solve such problems. This knowledge is important for anyone who wants to excel in this field. The basic of chemical equilibrium is the core of this discussion.

Understanding the Problem

Alright, here's what we've got. At 25°C, we have the reaction: HCl(g) ⇌ H₂(g) + Cl₂(g) with an equilibrium constant K = 4.17 × 10⁻³⁴. Our goal is to find the equilibrium constant (let's call it K') for the reaction: ½ H₂(g) + ½ Cl₂(g) ⇌ HCl(g). Notice a few key differences here? First, the original reaction is reversed. Second, the stoichiometric coefficients (the numbers in front of the chemical formulas) have been halved. This means the direction of the reaction is flipped, and the amount of reactants and products involved is changed. The equilibrium constant is a measure of the relative amounts of reactants and products present in a reaction at equilibrium. It tells us the extent to which a reaction will proceed to completion. Now that we understand the problem, we can continue to explain the process.

The Relationship Between K and K'

So, how do we relate the original K to the new K'? There are two main rules to remember:

Reversing the reaction: When you reverse a reaction, you take the inverse of the equilibrium constant. Mathematically, if you have A ⇌ B with K, then B ⇌ A has K' = 1/K.
Multiplying the coefficients: If you multiply all the coefficients in a balanced equation by a factor 'n', you raise the equilibrium constant to the power of 'n'. For example, if you have A ⇌ B with K, then nA ⇌ nB has K' = Kⁿ.

Let's apply these rules to our problem. We started with HCl(g) ⇌ H₂(g) + Cl₂(g) with K = 4.17 × 10⁻³⁴. Our target reaction is ½ H₂(g) + ½ Cl₂(g) ⇌ HCl(g).

Firstly, we need to reverse the original equation. Reversing it gives us H₂(g) + Cl₂(g) ⇌ HCl(g) with a new K value of K₁ = 1 / (4.17 × 10⁻³⁴).

Secondly, compare it to the target reaction, the coefficients has been multiplied by a factor of -1/2, so the K value must be raised by the same factor. So, K' = K₁^(1/2). Now we know the relationship between K and K', we can continue to solve the problem with the exact mathematical formulation.

Calculating K'

Now, let's crunch the numbers!

Reverse the reaction: K₁ = 1 / K = 1 / (4.17 × 10⁻³⁴) ≈ 2.398 × 10³³. This is the equilibrium constant for the reversed reaction, where HCl is the product.
Halve the coefficients: Because all coefficients in the new reaction is halved compared to the original reaction, the K value must be raised by the power of 1/2.

K' = K₁^(1/2) = (2.398 × 10³³)^(1/2) ≈ 4.897 × 10¹⁶

So, the equilibrium constant (K') for the reaction ½ H₂(g) + ½ Cl₂(g) ⇌ HCl(g) is approximately 4.897 × 10¹⁶.

Implications of the Result

Wow, that's a huge K'! Remember, a large K indicates that the products are strongly favored at equilibrium. In this case, the value of K' is significantly greater than 1, implying that the reaction will strongly favor the formation of HCl when the initial reactants are half a mole of H₂ and Cl₂ each. This makes perfect sense when you look at the reaction. Since the original reaction had a tiny K, it means the reverse reaction (forming HCl) is highly favorable. This illustrates a fundamental principle of chemical equilibrium: changes in reaction conditions (like reversing the reaction or changing the coefficients) dramatically impact the position of equilibrium and the value of K. So, always remember that, guys! The value of K tells you a lot about the reaction.

Summary and Key Takeaways

Let's recap what we've learned and some key things to remember:

Equilibrium Constant (K): This value describes the ratio of products to reactants at equilibrium. A large K means the products are favored.
Reversing a reaction: Take the inverse (1/K) of the original K.
Multiplying coefficients: Raise K to the power of the multiplier. For example, if the coefficients are doubled, square the K value (K²).
Putting it together: By understanding these two rules, you can calculate the equilibrium constant for modified reactions.

So, when you see a problem like this, remember these steps. First, identify the changes to the original reaction. Second, apply the appropriate rules to find the new K. Chemistry, like most subjects, builds on itself. This is a fundamental concept that you will use again and again. You will face problems like this often.

Practical Applications and Further Exploration

This principle is not just a theoretical concept. It has several practical applications. Consider industrial processes where you might want to optimize the yield of a product. By manipulating reaction conditions, such as temperature or pressure (though these don't directly change K, they can shift the equilibrium), you can affect the position of equilibrium and maximize product formation. This is super important in chemical engineering, where they're constantly trying to make processes more efficient. For example, you might want to make a particular reaction go towards a direction, such as producing a product. By manipulating reaction conditions, you can change the position of equilibrium and maximize the production.

If you want to dive deeper, you could explore how temperature affects K (Le Chatelier's Principle!), the concept of Gibbs Free Energy (which links K to spontaneity), or how to calculate K from initial concentrations and equilibrium concentrations. You can also work more practice problems to master this topic.

Conclusion

And there you have it! We've successfully calculated K' for the modified reaction. Remember the key rules: reverse the reaction (invert K), and adjust for coefficient changes (raise K to the power). Keep practicing these types of problems, and you'll become a pro in no time! Chemistry can seem complex at times, but by breaking things down and focusing on the core concepts, it becomes much more manageable. Keep exploring, keep questioning, and you'll find that chemistry is a fascinating field. Good luck, and keep up the great work, everyone!