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. (a) What fraction of the CCl 4 is converted into C and Cl 2 ?

Hello everyone. So in this video we're taking a look at our ice box, but we're gonna go ahead and apply this icebox to pressures instead of our traditional moles that were usually used to, so let's go ahead and rewrite our equation that's given to us in the problem, which is right over here. So we have our N 204 And then we have our equilibrium arrows and two moles of RNO. two. So of course, to set up the icebox, we need the I. C N E. If you call our I sense for initial we have received for change and then we have our E for equilibrium before I start. Let's go ahead. And also remind ourselves that on the left side we're gonna go ahead and lose reactant since and we're gonna go ahead and gain products. Okay, something to keep in mind when we're doing our filling up this ice table. Okay, so for our initial N204, that's going to be 750 tours For RN. 0. 2, we will have none for initial again, because we are losing reactant, we're gonna go ahead and minus X. Because we don't know how much exactly. And as for our end to we're gonna go ahead and gain the product. So we'll have plus two X. Because we see that too there. Alright, now at equilibrium then there'll be 7 50 minus X. For the equilibrium on the right side is going to be zero minus or zero plus R. Two X. And that's just going to be two X. So now that we have this table then we can go ahead and calculate for our constant value R K p value. The equation for RKP value is going to be the pressure of RN. 0 2 to the second power divided by P L'art N 204. So in the problem they gave us that K. P value. We go ahead and actually use that then. So we have 47. equal into and from the ice table and go ahead and fill out the pressures. So on the numerator will be two X to the second power and then we have 7 50 minus X in the denominator. So simplifying this just a little we get set 47. B. Going to four X squared all over again. 7 50 minus X. We go ahead and scroll down a little bit so we have more room. So sort of go ahead to isolate or just simplifying this here. Then our gets -47.9 X. Equalling 24 X squared. We're gonna go ahead and put everything on one side so we'll get zero equal into four X squared plus 47.9 X minus R. 35,925. And as you can see here then we have something that looks sort of familiar. We can actually go ahead and use the quadratic formula. So if we use our quadratic formula which is available on a calculator, we'll get that the X. value will equal to 89. Tour. So the pressure of N. 02 because you go to two X. That we said. So plugging our X. And then so 89 0. We'll get that. The P. Of N. 02 is equal to 178. Tour. I'll go ahead and start this over here now. So we know that P is proportional to most. So we can simply use the pressures to calculate the more fraction then so mole fraction of R. N. 02 is equal to moles of N. 02 over the total moles Which of course always says it is proportional will be pressure of n. 0. 2 over the total pressure. So plugging in those values then we'll have 178 tour that will just calculate for on top on the bottom. It'll be 178 tour Plus Our 7 50 -89. Alright, so putting that into my calculator, my mole fraction will equal to 0.212. It doesn't need darker colour. So 212. That will be my mole faction. Alright, this is going to be my final answer for this problem. Thank you all so much for watching

Ch.15 - Chemical Equilibrium

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Brown 15th Edition

Ch.15 - Chemical Equilibrium

Problem 89a

Chapter 15, Problem 89a

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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Start by writing the balanced chemical equation for the reaction: CCl4(g) ⇌ C(s) + 2 Cl2(g). Note that C(s) is a solid and does not appear in the expression for the equilibrium constant Kp.

Express the equilibrium constant Kp in terms of the partial pressures of the gases involved. Since Kp = 0.76, we have: Kp = (P_Cl2)^2 / P_CCl4.

Define the change in pressure for the reaction. Let x be the change in pressure of CCl4 that is converted to products. At equilibrium, the pressure of CCl4 will be (2.00 - x) atm, and the pressure of Cl2 will be 2x atm.

Substitute the equilibrium pressures into the expression for Kp: 0.76 = (2x)^2 / (2.00 - x).

Solve the equation for x to find the change in pressure. The fraction of CCl4 converted is x / 2.00. This will give you the fraction of the initial CCl4 that is converted to C and Cl2 at equilibrium.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (Kp)

The equilibrium constant (Kp) is a dimensionless value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. For gaseous reactions, Kp is calculated using partial pressures. In this case, Kp = 0.76 indicates the relationship between the partial pressures of CCl4, C, and Cl2 when the system reaches equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust to counteract the change and restore a new equilibrium. This principle helps predict how changes in concentration, pressure, or temperature will affect the position of equilibrium, which is essential for understanding how the reaction will shift when CCl4 is introduced into the flask.

Stoichiometry of the Reaction

Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. For the reaction CCl4(g) ⇌ C(s) + 2 Cl2(g), the stoichiometry indicates that one mole of CCl4 produces one mole of solid carbon and two moles of chlorine gas. Understanding stoichiometry is crucial for calculating the fraction of CCl4 converted to products at equilibrium based on the initial concentration and the equilibrium constant.

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Textbook Question

At 900 °C, 𝐾_𝑐= 0.0108 for the reaction

CaCO₃(𝑠) ⇌ CaO(𝑠) + CO₂(𝑔)

A mixture of CaCO₃, CaO, and CO₂ is placed in a 10.0-L vessel at 900°C. For the following mixtures, will the amount of CaCO3 increase, decrease, or remain the same as the system approaches equilibrium?

(a) 15.0 g CaCO3, 15.0 g CaO, and 4.25 g CO2

(b) 2.50 g CaCO3, 25.0 g CaO, and 5.66 g CO2

Textbook Question

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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Textbook Question

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?

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Textbook Question

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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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. (a) What fraction of the CCl4 is converted into C and Cl2?

Verified video answer for a similar problem:

Key Concepts

Equilibrium Constant (Kp)

Le Chatelier's Principle

Stoichiometry of the Reaction

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?