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Which of the following statements is correct about a reacting system at equilibrium: (a) the concentrations of reactants are equal to the concentrations of products, (b) the rate of the forward reaction is equal to the rate of the reverse reaction.

Short Answer

Expert verified
Statement (b) is correct; at equilibrium, reaction rates are equal.

Step by step solution

01

Understanding Equilibrium

At equilibrium, a chemical reaction occurs at a state where the rates of the forward and reverse reactions are equal. This means that the quantities of reactants and products remain constant over time, though not necessarily equal to each other.
02

Analyzing Statement (a)

Statement (a) claims that at equilibrium, the concentrations of reactants are equal to the concentrations of products. This is generally incorrect. Equilibrium does not require the concentrations to be equal, but rather that they remain constant over time.
03

Analyzing Statement (b)

Statement (b) states that at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This is indeed a correct statement, as the hallmark of equilibrium is the equality of the reaction rates in both directions.
04

Conclusion

Based on the understanding of chemical equilibrium, statement (b) is correct because the definition of equilibrium is characterized by equal reaction rates. Statement (a) is incorrect as it misinterprets the concept by equating the concentrations rather than the rates.

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Key Concepts

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

Reaction Rates
In a chemical reaction, the rate signifies how fast reactants transform into products. The concept of reaction rates is crucial in understanding chemical equilibrium. When a reaction is taking place, the rate at which reactants get used up and products form is initially swift. Over time, as products accumulate and reactants deplete, these rates slow down until equilibrium is reached.

Imagine a scenario where a group of people are running a race. At the start, they are full of energy and move quickly, but as they near the finish line, their pace slows and becomes steady. This is similar to how reaction rates behave as they approach equilibrium. At this point, neither the reactants nor the products 'win' in terms of concentration change; they simply reach a standstill where their rates are balanced.

Understanding the balance and speed of reaction rates is key to comprehending how chemical equilibrium works and why reactions progress.
  • Initially fast but slows over time
  • Equilibrium is where reaction rates equalize
  • Rates are influenced by factors like concentration, temperature, and presence of catalysts
Forward and Reverse Reactions
When a chemical reaction occurs, two competing processes take place: the forward and reverse reactions. The forward reaction is where reactants form products. Conversely, the reverse reaction is where products revert back into reactants. At equilibrium, these two reactions happen simultaneously and at the same rate.

This delicate balance is much like a busy street with cars traveling back and forth at a steady pace; there is motion, yet things remain orderly and ongoing. Both forward and reverse reactions are essential for maintaining the stability of a chemical system at equilibrium.
The concept of forward and reverse reactions helps to explain why concentrations stay constant over time at equilibrium, even though reactions continue.
  • Forward and reverse reactions occur simultaneously
  • At equilibrium, rates of these reactions are equal
  • Both reactions are essential for the dynamic balance
Concentrations of Reactants and Products
The concentration of reactants and products is fundamental to understanding chemical equilibrium. In the context of equilibrium, concentrations do not necessarily have to be equal, but they do become constant over time.

Think of it like water in a bathtub where the faucet and the drain are open, but the water level remains constant because the inflow from the faucet matches the outflow from the drain. Similarly, at equilibrium, the concentrations of reactants and products reach a steady state.

It is important to distinguish that while concentrations stabilize, it doesn't imply they are identical, just that they are fixed due to the balancing act of the forward and reverse reactions. This stability in concentration is pivotal since it allows chemists to predict how a reaction mixture will behave over time.
  • Concentrations become constant at equilibrium
  • Equal concentrations are not a requirement
  • Stability is achieved through balanced reaction rates

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Most popular questions from this chapter

When heated at high temperatures, iodine vapor dissociates as follows: $$ \mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{I}(g) $$ In one experiment, a chemist finds that when 0.054 mole of \(\mathrm{I}_{2}\) was placed in a flask of volume \(0.48 \mathrm{~L}\) at \(587 \mathrm{~K},\) the degree of dissociation (i.e., the fraction of \(\mathrm{I}_{2}\) dissociated) was \(0.0252 .\) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the reaction at this temperature.

One mole of \(\mathrm{N}_{2}\) and three moles of \(\mathrm{H}_{2}\) are placed in a flask at \(375^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is 0.21 . The \(K_{p}\) for the reaction is \(4.31 \times 10^{-4}\).

A mixture of 0.47 mole of \(\mathrm{H}_{2}\) and 3.59 moles of \(\mathrm{HCl}\) is heated to \(2800^{\circ} \mathrm{C}\). Calculate the equilibrium partial pressures of \(\mathrm{H}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{HCl}\) if the total pressure is 2.00 atm. For the reaction: $$ \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{HCl}(g) $$ \(K_{P}\) is 193 at \(2800^{\circ} \mathrm{C}\).

About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=206 \mathrm{~kJ} / \mathrm{mol}\) The secondary stage is carried out at about \(1000^{\circ} \mathrm{C},\) in the presence of air, to convert the remaining methane to hydrogen: \(\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftarrows \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \quad \Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol}\) (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stages? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}:\) $$\mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftarrows \mathrm{NaCl}(l)+\mathrm{K}(g)$$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C}\), respectively.)

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