Explain what is wrong with the following statements. (a) Once a reaction has reached equilibrium, all reaction stops. (b) If more reactant is used, the equilibrium constant will have a larger value.

Short Answer

Expert verified
Statement (a) is incorrect because at equilibrium, reactions do not stop but continue occurring with no net change in concentrations. Statement (b) is also incorrect since the equilibrium constant is unaffected by the initial amounts of reactants.

Step by step solution

01

Analyzing Statement (a)

Understand the concept of chemical equilibrium. When a reaction reaches equilibrium, it means that the rate of the forward reaction is equal to the rate of the reverse reaction. The concentrations of the reactants and products remain constant over time, but the reactions continue to occur at a microscopic level.
02

Explaining the Misconception in Statement (a)

Clarify that the statement is incorrect because reactions still take place even after equilibrium has been established. Although there are no net changes in the concentrations of the reactants and products, both the forward and reverse reactions continue to occur.
03

Analyzing Statement (b)

Review the definition of the equilibrium constant (K). The equilibrium constant is a ratio of the concentration of products to reactants at equilibrium, raised to the power of their coefficients in the balanced chemical equation. It is a measure of the extent of the reaction and is not dependent on the initial concentrations of reactants or products.
04

Explaining the Misconception in Statement (b)

Point out that the equilibrium constant is a fixed value at a given temperature and does not change with the amounts of reactants used. Adding more reactant may shift the position of the equilibrium, affecting the concentrations of reactants and products at equilibrium, but it does not change the value of the equilibrium constant itself.

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Key Concepts

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

Equilibrium Constant
In the world of chemistry, the equilibrium constant is a vital number that tells us a lot about a chemical reaction. It represents the balance point of a reaction when the rate of the forward reaction equals the rate of the reverse reaction. Let's imagine a simple seesaw with products on one side and reactants on the other. The equilibrium constant, denoted as K, is like the point where the seesaw doesn't tilt either way, it's perfectly horizontal.

Now, let's clear up a common mistake: some think that if you pile on more reactants onto our imaginary seesaw, K will change. But that's not quite right! K stays the same, no matter how much reactant you add. Why? Because it's determined by the ratio of the products to the reactants at equilibrium — think of it like a recipe that always needs the same proportions of ingredients to taste just right, no matter how much you make.

Here's the equation for the equilibrium constant:
\( K = \frac{[products]}{[reactants]} \)
Where the square brackets represent the concentration of substances and we raise them to the power of their coefficients from the balanced chemical equation. The key takeaway: K is consistent for a given reaction at a certain temperature, and does not change with the amount of reactants or products you start with.
Reaction Rates
Let's talk about reaction rates in a way that would help anyone understand, like comparing it to a dance-off between two teams. In chemistry, when we say 'reaction rates,' we are talking about the speed at which the reactant molecules dance (react) to form product molecules. When a reaction starts, the teams are all hyped up and the dance-off (reaction) is super-fast. But as time goes on, the teams get tired (the concentration of reactants decreases) and the dance-off slows down.

At equilibrium, something neat happens — the speed of the forward team (forward reaction) matches that of the reverse team (reverse reaction). It's like both teams are matching each other's moves perfectly. This doesn't mean they’ve stopped dancing! They're still grooving, but now the score stays even. So, even when it looks like things are still, there's a whole bunch of action going on at a microscopic level.

To measure the speed of this chemistry dance, scientists use rates. The faster the reaction, the higher the rate. But remember, at equilibrium, the rates of the forward and reverse reactions are the same, and that special point is what makes chemical equilibrium so fascinating and important in understanding how reactions work.
Balanced Chemical Equation
What if we think of a balanced chemical equation like a perfectly portioned recipe? In cooking, if you don't balance your ingredients, your recipe doesn't turn out right. It's similar in chemistry. A balanced chemical equation ensures that we have the correct amount of each molecule dancing in our reaction so that none are left out and no one crashes the party with too much of themselves.

The law of conservation of mass tells us that we can't create or destroy atoms in a chemical reaction, so we have to balance the equation to show that. In other words, we're arranging the number of atoms of each element to be the same on both sides of the equation. This gives us a clear blueprint for understanding how much of a reactant we’ll need to produce a certain amount of product.

Here’s a simple example:
\[ \ce{2H_2 + O_2 -> 2H_2O} \]
This equation tells us that two molecules of hydrogen gas react with one molecule of oxygen gas to make two molecules of water. Easy, right? Just remember, like a perfect recipe, the numbers in front of each molecule (called coefficients) ensure that atoms are conserved and the dance of a chemical reaction can happen smoothly.

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Most popular questions from this chapter

A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

(a) Calculate the reaction free energy of \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Hl}(\mathrm{g})\) at \(700 \mathrm{~K}\) when the concentrations of \(\mathrm{H}_{2}, \mathrm{I}_{2}\), and HI are \(0.026,0.33\), and \(1.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), respectively. For this reaction, \(K_{c}=54\) at \(700 \mathrm{~K}\). (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

At \(1565 \mathrm{~K}\), the equilibrium constants for the reactions (1) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) and \((2)\) \(2 \mathrm{CO}_{2}(\mathrm{~g})=2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) are \(1.6 \times 10^{-11}\) and \(1.3 \times 10^{-10}\), respectively. (a) What is the equilibrium constant for the reaction (3) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})=\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\) at that temperature? (b) Show that the manner in which equilibrium constants are calculated is consistent with the manner in which the \(\Delta G_{1}^{\circ}\) values are calculated when combining two or more equarions by determining \(\Delta G_{e}{ }^{\circ}\) for \((1)\) and \((2)\) and using those values to calculare \(\Delta G,{ }^{\circ}\) and \(K_{3}\) for reaction (3).

A mixture of \(0.0560 \mathrm{~mol} \mathrm{O}_{2}\) and \(0.0200 \mathrm{~mol}\) \(\mathrm{N}_{2} \mathrm{O}\) is placed in a \(1.00-\mathrm{L}\) reaction vessel at \(25^{\circ} \mathrm{C}\). When the reaction \(2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{NO}_{2}(\mathrm{~g})\) is at equilibrium, \(0.0200\) mol \(\mathrm{NO}_{2}\) is present. (a) What are the equilibrium concentrations? (b) What is the value of \(K_{c}\) ?

Use Le Chatelier's principle to predict the effect that the change given in the first column of the following table has on the quantity in the second column for the following equilibrium system: $$ \begin{gathered} 5 \mathrm{CO}(\mathrm{g})+\mathrm{I}_{2} \mathrm{O}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{I}_{2}(\mathrm{~g})+5 \mathrm{CO}_{2}(\mathrm{~g}) \\\ \Delta H^{\circ}=-1175 \mathrm{~kJ} \end{gathered} $$

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