At 700 K, the equilibrium constant for the reaction CCl 4 (𝑔) ⇌ C(𝑠) + 2 Cl 2 (𝑔) is 𝐾 𝑝 = 0.76. A flask is charged with 2.00 atm of CCl 4 , which then reaches equilibrium at 700 K. (b) What are the partial pressures of CCl 4 and Cl 2 at equilibrium?

Step 1: Write the balanced chemical equation for the reaction: CCl_{4}(g) ⇌ C(s) + 2 Cl_{2}(g).. Step 2: Define the initial pressures of the reactants and products. Initially, the pressure of CCl_{4} is 2.00 atm, and the pressures of C(s) and Cl_{2} are 0 atm.. Step 3: Let the change in pressure of CCl_{4} at equilibrium be -x atm. Since C(s) is a solid, its pressure does not contribute to the equilibrium expression. The change in pressure of Cl_{2} will be +2x atm.. Step 4: Write the expression for the equilibrium constant, K_{p}, in terms of the partial pressures of the gases at equilibrium. K_{p} = (P_{Cl2}$$)^2 / P$$_{CCl4}.. Step 5: Substitute the expressions for the pressures at equilibrium into the K_{p} expression and solve for x. Use the quadratic formula if necessary to find the value of x, and then calculate the equilibrium partial pressures of CCl_{4} and Cl_{2}.

At 700 K, the equilibrium constant for the reaction CCl4(𝑔) ⇌ C(𝑠) + 2 Cl2(𝑔) is 𝐾𝑝 = 0.76. A flask is charged with 2.00 atm of CCl4, which then reaches equilibrium at 700 K. (b) What are the partial pressures of CCl4 and Cl2 at equilibrium?

Step 1: Write the balanced chemical equation for the reaction: CCl_{4}(g) ⇌ C(s) + 2 Cl_{2}(g). Step 2: Define the initial pressures of the reactants and products. Initially, the pressure of CCl_{4} is 2.00 atm, and the pressures of C(s) and Cl_{2} are 0 atm. Step 3: Let the change in pressure of CCl_{4} at equilibrium be -x atm. Since C(s) is a solid, its pressure does not contribute to the equilibrium expression. The change in pressure of Cl_{2} will be +2x atm. Step 4: Write the expression for the equilibrium constant, K_{p}, in terms of the partial pressures of the gases at equilibrium. K_{p} = (P_{Cl2}$$)^2 / P$$_{CCl4}. Step 5: Substitute the expressions for the pressures at equilibrium into the K_{p} expression and solve for x. Use the quadratic formula if necessary to find the value of x, and then calculate the equilibrium partial pressures of CCl_{4} and Cl_{2}.

Ch.15 - Chemical Equilibrium

Brown - Chemistry: The Central Science 15th Edition

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Brown 15th Edition

Ch.15 - Chemical Equilibrium

Problema 89b

Capitolo 15, Problema 89b

At 700 K, the equilibrium constant for the reaction CCl₄(𝑔) ⇌ C(𝑠) + 2 Cl₂(𝑔) is 𝐾_𝑝= 0.76. A flask is charged with 2.00 atm of CCl₄, which then reaches equilibrium at 700 K. (b) What are the partial pressures of CCl₄ and Cl₂ at equilibrium?

Guida verificata passo dopo passo

Step 1: Write the balanced chemical equation for the reaction: CCl₄(g) ⇌ C(s) + 2 Cl₂(g).

Step 2: Define the initial pressures of the reactants and products. Initially, the pressure of CCl₄ is 2.00 atm, and the pressures of C(s) and Cl₂ are 0 atm.

Step 3: Let the change in pressure of CCl₄ at equilibrium be -x atm. Since C(s) is a solid, its pressure does not contribute to the equilibrium expression. The change in pressure of Cl₂ will be +2x atm.

Step 4: Write the expression for the equilibrium constant, K_p, in terms of the partial pressures of the gases at equilibrium. K_p = (P_Cl₂)^2 / P_CCl₄.

Step 5: Substitute the expressions for the pressures at equilibrium into the K_p expression and solve for x. Use the quadratic formula if necessary to find the value of x, and then calculate the equilibrium partial pressures of CCl₄ and Cl₂.

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Equilibrium Constant (Kp)

The equilibrium constant (Kp) is a numerical value that expresses the ratio of the partial pressures of the products to the reactants at equilibrium for a given reaction at a specific temperature. For the reaction CCl₄(g) ⇌ C(s) + 2 Cl₂(g), Kp = 0.76 indicates that at equilibrium, the ratio of the pressure of Cl₂ squared to the pressure of CCl₄ is constant. This concept is crucial for determining the concentrations or pressures of reactants and products at equilibrium.

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Equilibrium Constant Expressions

Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium is disturbed by a change in concentration, temperature, or pressure, the system will adjust to counteract the disturbance and restore a new equilibrium. In this case, if the initial pressure of CCl₄ is altered, the system will shift to either produce more products or reactants to re-establish equilibrium, which is essential for calculating the final partial pressures.

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Le Chatelier's Principle

Partial Pressure

Partial pressure is the pressure exerted by a single component of a gas mixture. According to Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressures of each individual gas. In the context of the given reaction, understanding how to calculate the partial pressures of CCl₄ and Cl₂ at equilibrium is necessary to solve the problem, as it involves using the initial conditions and the equilibrium constant.

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Partial Pressure Calculation

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Domanda del libro di testo

At 700 K, the equilibrium constant for the reaction CCl₄(𝑔) ⇌ C(𝑠) + 2 Cl₂(𝑔) is 𝐾_𝑝= 0.76. A flask is charged with 2.00 atm of CCl₄, which then reaches equilibrium at 700 K. (a) What fraction of the CCl₄ is converted into C and Cl₂?

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At a temperature of 700 K, the forward and reverse rate constants for the reaction 2 HI(g) ⇌ H₂(g) + I₂(g) are k_f = 1.8×10⁻³⁰ M⁻¹s⁻¹ and k_r = 0.063 M⁻¹s⁻¹.

(a) What is the value of the equilibrium constant K_c at 700 K?

(b) Is the forward reaction endothermic or exothermic if the rate constants for the same reaction have values of k_f= 0.097M⁻¹s⁻¹ and k_r= 2.6 M⁻¹s⁻¹ at 800 K?

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The equilibrium constant constant 𝐾_𝑐 for C(𝑠) + CO₂(𝑔) ⇌ 2 CO(𝑔) is 1.9 at 1000 K and 0.133 at 298 K. (a) If excess C is allowed to react with 25.0 g of CO2 in a 3.00-L vessel at 1000 K, how many grams of CO are produced? (b) If excess C is allowed to react with 25.0 g of CO₂ in a 3.00-L vessel at 1000 K, how many grams of C are consumed?

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Consider the hypothetical reaction A(𝑔) + 2 B(𝑔) ⇌ 2 C(𝑔), for which 𝐾_𝑐= 0.25 at a certain temperature. A 1.00-L reaction vessel is loaded with 1.00 mol of compound C, which is allowed to reach equilibrium. Let the variable x represent the number of mol/L of compound A present at equilibrium.

(d) The equation from part (c) is a cubic equation (one that has the form ax3 + bx2 + cx + d = 0). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of x that you specified in part (b). The point at which the cubic equation crosses the x-axis is the solution.

(e) From the plot in part (d), estimate the equilibrium concentrations of A, B, and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

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