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If \(Q=1.0 \times 10^{40}\) for the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\), will the reaction have a tendency to form products or reactants or will it be at equilibrium?

Short Answer

Expert verified
Without knowing the equilibrium constant (K), we cannot say definitively whether the reaction will form more products or reactants, but a very large Q value suggests a tendency to form reactants.

Step by step solution

01

Understand the concept of the Reaction Quotient (Q)

The Reaction Quotient (Q) is used to determine the direction in which a reaction will proceed to reach equilibrium. It is the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients, at a given moment in time. For gases, concentration is often replaced by partial pressures.
02

Compare the Reaction Quotient (Q) to the Equilibrium Constant (K)

To determine if the reaction will form more products or reactants, or if it's at equilibrium, compare the value of Q with the equilibrium constant (K) for the reaction at the given temperature (25°C). If Q < K, the reaction tends to form more products, moving towards equilibrium. If Q > K, the reaction tends to form more reactants. If Q = K, the reaction is at equilibrium.
03

Analyze the given Q value

Since the value of Q is given as 1.0 x 10^{40}, without the value of K we can't definitively say what direction the reaction will go. However, a Q value that high suggests that the reaction may have proceeded far to the right, producing a large amount of products.
04

Infer the tendency of the reaction

Although we cannot make a definite inference without knowing the equilibrium constant (K), a very large Q value often indicates that the reaction quotient is greater than K, and therefore the system would likely shift to the left, favoring the formation of reactants.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is crucial in predicting the behavior of a chemical system over time. It refers to a state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time.

At equilibrium, the system is stable, and the concentrations of all reactants and products remain constant, though not necessarily equal. This balance does not mean that the reaction has stopped; instead, both forward and reverse reactions are still occurring but at the same speed. In a closed system, once equilibrium is reached, it can be disturbed by changing conditions such as temperature, pressure, or concentration, leading to a shift in the equilibrium to re-establish balance, according to Le Châtelier's principle.
Equilibrium Constant (K)
The equilibrium constant (K) is a numerical value that reflects the ratio of product concentrations to reactant concentrations for a reaction at equilibrium, with each concentration raised to the power of its stoichiometric coefficient.

The value of K provides insights into the position of equilibrium: a large K (much greater than 1) indicates that the products are favored under equilibrium conditions, whereas a small K value (much less than 1) implies that reactants are favored. Importantly, K is temperature dependent and remains constant only at a given temperature. If the temperature changes, so does the value of K. To apply the concept of K effectively, it is essential to know the balanced chemical equation and understand that solid and pure liquid concentrations are not included in the calculation as they are considered to have constant values.
Chemical Reactions
Chemical reactions are processes where substances, known as reactants, undergo a transformation to form new substances called products. Depending on the nature of the products and reactants, these reactions can be classified into several types, such as synthesis, decomposition, single replacement, double replacement, and combustion reactions.

The course of a chemical reaction often involves breaking bonds in the reactants and forming new bonds in the products. This process is guided by factors like temperature, concentration, pressure, and catalysts that can influence the rate and extent of a reaction. A fundamental understanding of chemical reactions allows chemists to predict the outcome of a reaction, influence reaction conditions to favor the desired product, or synthesize new compounds with specific properties.

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Most popular questions from this chapter

State what happens to the concentration of the indicated substance when the total pressure on each of the following equilibria is increased (by compression): (a) \(\mathrm{NO}_{2}\) (g) in \(2 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{PbO}\) (s) \(+4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}\) (g) (b) \(\mathrm{NO}(\mathrm{g})\) in \(3 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})\) (c) \(\mathrm{HI}(\mathrm{g})\) in \(2 \mathrm{HCl}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (d) \(\mathrm{SO}_{2}\) (g) in \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) (e) \(\mathrm{NO}_{2}\) (g) in \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\)

Determine whether the following statements are true or false. If false, explain why. (a) In an equilibrium reaction, the reverse reaction begins only when all reactants have been converted to products. (b) The equilibrium concentrations will be the same whether one starts with pure reactants or pure products. (c) The rates of the forward and reverse reactions are the same at equilibrium. (d) If the Gibbs free energy is greater than the standard Gibbs free energy of reaction, the reaction proceeds forward to equilibrium.

Determine \(K_{c}\) for each of the following equilibria from the value of \(K\) : (a) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.4\) at \(1000 . \mathrm{K}\) (b) \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}), K=9.4 \times 10^{-2}\) at \(24^{\circ} \mathrm{C}\)

Cyclohexane \((C)\) and methylcyclopentane \((M)\) are isomers with the chemical formula \(\mathrm{C}_{6} \mathrm{H}_{12}\). The equilibrium constant for the rearrangement \(\mathrm{C} \rightleftharpoons \mathrm{M}\) in solution is \(0.140\) at \(25^{\circ} \mathrm{C}\). (a) \(\mathrm{A}\) solution of \(0.0200 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) cyclohexane and \(0.100 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) methylcyclopentane is prepared. Is the system at equilibrium? If not, will it will form more reactants or more products? (b) What are the concentrations of cyclohexane and methylcyclohexane at equilibrium? (c) If the temperature is raised to \(50 .{ }^{\circ} \mathrm{C}\), the concentration of cyclohexane becomes \(0.100 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) when equilibrium is re-established. Calculate the new equilibrium constant. (d) Is the reaction exothermic or endothermic at \(25^{\circ} \mathrm{C}\) ? Explain your conclusion.

The equilibrium constant for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) has the value \(K=2.5 \times 10^{10}\) at 500 . \(\mathrm{K}\). Find the value of \(K\) for each of the following reactions at the same temperature. (a) \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\) (b) \(\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) (c) \(3 \mathrm{SO}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{SO}_{3}(\mathrm{~g})\)

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