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(a) If \(Q_{c}>K_{c}\) how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

Short Answer

Expert verified
(a) If \(Q_c > K_c\), the reaction must shift in the reverse direction to reach equilibrium. (b) At the start of the reaction when only reactants are present and no products have been formed, the value of \(Q_c = 0\).

Step by step solution

01

(a) Understanding the relationship between Qc and Kc

The Reaction Quotient (\(Q_c\)) is a measure of the progress of a reaction, while the Equilibrium Constant (\(K_c\)) tells us the ratio of products to reactants when the reaction is at equilibrium. When \(Q_c = K_c\), the reaction is at equilibrium. If \(Q_c > K_c\), there are more products than reactants compared to the equilibrium state. Conversely, if \(Q_c < K_c\), there are more reactants than products compared to the equilibrium state.
02

(a) Determining the direction of the reaction

In this exercise, we are given that \(Q_c > K_c\). This means that there are more products than reactants compared to the equilibrium state. To reach equilibrium, the reaction must proceed in the direction that will decrease the amount of products and increase the amount of reactants. In other words, the reaction must shift in the reverse direction.
03

(b) Identifying the starting conditions

At the beginning of the reaction, only reactants are present, and no products have been formed. Since \(Q_c\) is calculated as the ratio of concentrations of products to reactants, we can write the general expression for \(Q_c\) as: \[Q_c = \frac{[Product_1]^{x_1}[Product_2]^{x_2}\cdots}{[Reactant_1]^{y_1}[Reactant_2]^{y_2}\cdots}\] where \(x_i\) and \(y_i\) are the stoichiometric coefficients of the respective products and reactants.
04

(b) Calculate Qc at the beginning of the reaction

Since no products have been formed, the concentration of all products is zero at the beginning of the reaction. Therefore, the numerator of the \(Q_c\) expression will be a product of zeros. Since any number multiplied by zero is zero, the entire numerator becomes zero: \[Q_c = \frac{0}{[Reactant_1]^{y_1}[Reactant_2]^{y_2}\cdots}\] The value of \(Q_c\) at this point in the reaction is therefore \(Q_c = 0\).

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Key Concepts

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

Equilibrium Constant (Kc)
The equilibrium constant, denoted as \( K_c \), plays a crucial role in understanding chemical reactions. It is a number that expresses the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium, each raised to the power of their respective stoichiometric coefficients in the balanced equation. Mathematically, it is represented as:
\[ K_c = \frac{[Product_1]^{coeff_{p1}}[Product_2]^{coeff_{p2}}\ldots}{[Reactant_1]^{coeff_{r1}}[Reactant_2]^{coeff_{r2}}\ldots} \]
where the square brackets denote concentrations, and \( coeff_\{pi\} \) and \( coeff_\{ri\} \) are the coefficients of the products and reactants, respectively.

Importance of \( K_c \)

Knowing the value of \( K_c \) gives insight into the position of equilibrium. A high \( K_c \) value indicates a tendency for the reaction to yield more products, whereas a low \( K_c \) suggests a reaction favoring the reactants. It is pivotal to remember that equilibrium does not mean the concentrations of reactants and products are equal, but rather that their rates of formation and consumption are balanced, resulting in constant concentrations. Temperature affects \( K_c \) significantly; as the conditions change, so does the value of the equilibrium constant, reflecting a shift in the balance point of the reaction.
Chemical Equilibrium
Chemical equilibrium is the state of a reaction in which the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time. This does not mean the reaction has stopped; rather, the dynamic process is ongoing, but the opposing reactions balance each other out.

Dynamic Nature of Equilibrium

Even at equilibrium, the reactant molecules are continuously converting into product molecules and vice versa. The dynamic nature of equilibrium implies that while the macroscopic properties remain unchanged, microscopic changes are always happening within the system.
Knowing when a system has reached chemical equilibrium is essential for predicting the outcome of reactions and optimizing conditions for the desired product. To quantify this state, chemists use the equilibrium constant (as discussed above), which remains unchanged at a given temperature as long as the system remains undisturbed.
Reaction Direction
Determining the reaction direction is key to predicting how a reaction will proceed to reach equilibrium. The reaction quotient, \( Q_c \), helps in predicting this direction before the system has reached equilibrium.

Using \( Q_c \) to Predict Reaction Direction

When a reaction has not yet reached equilibrium, comparing the reaction quotient (\( Q_c \)) to the equilibrium constant (\( K_c \)) informs us whether the reaction will proceed forward or reverse to reach equilibrium:
  • If \( Q_c < K_c \), the reaction will shift forward, favoring the formation of products.
  • If \( Q_c > K_c \), the reaction will shift in the reverse direction, favoring the formation of reactants.
This can be exemplified with Le Chatelier's principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will move to counteract the change. Factors such as concentration, pressure, and temperature changes can influence the reaction direction. Understanding these principles is essential for controlling chemical processes and yields. As an example, in the context of the exercise, the initial value of \( Q_c \) being zero indicates that the reaction will proceed in the forward direction to begin producing products and eventually establish equilibrium.

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Most popular questions from this chapter

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons\) \(\mathrm{PCl}_{5}(g) . \mathrm{A} 7.5-\mathrm{L}\) gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g),\) which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are \(P_{\mathrm{PCl}_{3}}=0.124 \mathrm{atm}, P_{\mathrm{Cl}_{2}}=0.157 \mathrm{atm},\) and \(P_{\mathrm{PCl}_{\mathrm{s}}}=1.30 \mathrm{atm}\) (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at 450 \(\mathrm{K}\)

For the reaction \(\mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{IBr}(g), K_{c}=280\) at \(150^{\circ} \mathrm{C} .\) Suppose that 0.500 \(\mathrm{mol} \mathrm{IBr}\) in a 2.00 -L flask is allowed to reach equilibrium at \(150^{\circ} \mathrm{C}\) . What are the equilibrium concentrations of IBr, I \(_{2},\) and \(\mathrm{Br}_{2} ?\)

Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24^{\circ} \mathrm{C} .\) The following reaction takes place: $$\mathrm{NH}_{4} \mathrm{SH}(s) \Longrightarrow \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ At equilibrium, the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) taken together) is 0.614 atm. What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C} ?\)

A mixture of 0.2000 mol of \(\mathrm{CO}_{2}, 0.1000 \mathrm{mol}\) of \(\mathrm{H}_{2},\) and 0.1600 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 2.000 -L vessel. The following equilibrium is established at \(500 \mathrm{K} :\) $$\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (a) Calculate the initial partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2,}\) and \(\mathrm{H}_{2} \mathrm{O}\) (b) At equilibrium \(P_{\mathrm{H}_{2} \mathrm{O}}\) \(=3.51\) atm. Calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2},\) and \(\mathrm{CO} .(\mathbf{c})\) Calculate \(K_{p}\) for the reaction. (d) Calculate \(K_{c}\) for the reaction.

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at 700 \(\mathrm{K} .\) In a \(2.00-\mathrm{L}\) flask containing an equilibrium mixture of the three gases, there are 0.056 \(\mathrm{g} \mathrm{H}_{2}\) and 4.36 \(\mathrm{g} \mathrm{I}_{2}\) . What is the mass of HI in the flask?

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