Problem 5 What happens to the value of the... [FREE SOLUTION] (2024)

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Chapter 16: Problem 5

What happens to the value of the equilibrium constant for a reaction if thereaction equation is reversed? Multiplied by a constant?

Short Answer

Expert verified

Reversing a reaction results in the equilibrium constant being the reciprocal of the original. Multiplying the reaction by a constant raises the equilibrium constant to the power of that constant.

Step by step solution

01

Understanding Equilibrium Constant Reversal

The equilibrium constant for a reaction is denoted by \( K \). When a reaction is reversed, the equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant. So, if the forward reaction has an equilibrium constant \( K \), then the reverse reaction's equilibrium constant becomes \( \frac{1}{K} \).

02

Understanding Equilibrium Constant Multiplication

When the coefficients in a balanced chemical equation are multiplied by a constant, the equilibrium constant for the new reaction is the original equilibrium constant raised to the power of that constant. For instance, if the balanced equation is multiplied by a factor of \( n \), and the equilibrium constant for the original reaction is \( K \), then the equilibrium constant for the modified reaction is \( K^n \).

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium

Imagine a seesaw perfectly balanced with equal weights on both sides—that's like chemical equilibrium, but with reactants and products in a chemical reaction. It's the state where the rate of the forward reaction equals the rate of the reverse reaction, so the concentrations of reactants and products remain constant over time. Imagine you've got a reaction where A turns into B. At equilibrium, A is becoming B at the same speed B is turning back into A. There's a special number called the equilibrium constant, denoted as 'K', that tells us the ratio of these concentrations at equilibrium.

This constant is crucial as it can hint at the prevalence of reactants or products in the balanced equation. If K is large, products rule the playground; if K is small, reactants are in charge. In the exercise we're looking at, the student needs to understand that flipping the equation or changing the amounts of reactants or products will lead to a game of equilibrium constant manipulation.

Le Chatelier's Principle

Ever heard of a chemical's version of stubbornness? That's what Le Chatelier’s principle is all about. This principle is like the universe's way of saying 'I don't like change.' It predicts how a system at equilibrium will adjust when it's messed with. Say you're chilling at the beach, perfectly comfy, but then the sun moves and you're now in the shade. You adjust to get back in the sun, right? Similarly, when the equilibrium of a reaction is disturbed, say by adding more reactant or removing some product, the reaction will shift to counteract the change and restore the balance.

This adjustment could mean making more products or more reactants, depending on what's been tweaked. It's like a tug-of-war where each side—reactants and products—tries to maintain equilibrium in face of changes in concentration, temperature, or pressure.

Reaction Quotient

Think of the reaction quotient, 'Q', as a spy on a secret mission, sneaking around to figure out if a reaction is at equilibrium. It’s like the equilibrium constant's twin, except it can be calculated at any point in the reaction—not just at equilibrium. Here's the kicker: by comparing Q to K, you can predict which way the reaction will shift to reach equilibrium.

If Q is less than K, the reaction goes forward to make more products. If Q is more than K, the reaction says 'whoa, too much!' and goes in reverse to make more reactants. If Q equals K, the reaction is already kicking back at equilibrium, and no more shifting will occur. Our students learn to calculate this Q to anticipate the actions of the reaction under non-equilibrium conditions.

Equilibrium Constant Manipulation

Equilibrium constant manipulation isn’t about pulling rabbits out of hats, but it's still pretty magical for chemists. It's tweaking the balance of a chemical equation and observing the impact on the equilibrium constant, 'K'. As described in the exercise, flipping the chemical seesaw—reversing the reaction—means you flip K as well, taking its reciprocal.

However, when you multiply the coefficients by a number, say doubling the ingredients in a cake recipe, you're not just doubling 'K'. Instead, you raise it to the power of that number. If you triple the coefficients, K goes cubed. Understanding this helps students solve the puzzles of what-if scenarios, like 'What happens if...' in a chemical equation. It's these manipulations that lay down the foundations for predicting how concentrations affect the behavior of reactions in real laboratory conditions.

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Problem 5 What happens to the value of the... [FREE SOLUTION] (3)

Most popular questions from this chapter

Each reaction is allowed to come to equilibrium, and then the volume ischanged as indicated. Predict the effect (shift right, shift left, or noeffect) of the indicated volume change. MissED THis? Read Section 16.9 ; WatchKCV 16.9 a. \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{l}(\mathrm{s})\) (volume isincreased) b. \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})=2\mathrm{H}_{2}(\mathrm{~s})+\mathrm{S}_{2}(\mathrm{~g})\) (volume is decreased) c. \(\mathrm{I}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\rightleftharpoons 2 \mathrm{ICl}(\mathrm{g})\) (volume is decreased)Consider the reaction and the associated equilibrium constant: $$ a A(g)+l B(g) \Longrightarrow c C(g) \quad K_{c}-5.0 $$ And the equilibrium concentrations of \(A, B,\) and \(C\) for each value of \(a,b,\) and \(c\). Assume that the initial concentrations of \(A\) and \(B\) are each\(1.0 \mathrm{M}\) and that no product is present at the begin. ning of thereaction. a. \(a-1 ; b-1 ; c-2\) b. \(a-1 ; b-1 ; c-1\) c. \(a-2 ; b-1 ; c=1\) (set up equation for \(x ;\) don't solve)What happens to a chemical system at equilibrium when that cqullibrium isdisturbed?What is the significance of the equilibrium constant? What does a largeequilibrium constant tell us about a reaction? A small one?What is the definition of the reaction quotient \((Q)\) for a reaction? Whatdoes Q measure?
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Problem 5 What happens to the value of the... [FREE SOLUTION] (2024)
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