The equilibrium constant for the dissociation of molecular iodine, I 2 (g) ⇌ 2 I(g), at 800 K is K c = 3.1 × 10 –5 . (a) Which species predominates at equilibrium I 2 or I?

Hello. Everyone in this video trying to calculate for the equilibrium concentration of N. 02. It's going to be our product in our case. So first things first, we're gonna go ahead and write out what were given to us in this problem where were given to us is going to be the equilibrium constant. That's equal to 0.212 and that's at degrees Celsius. So this equilibrium constant value of 0.212. This value is just around one. So neither of the gasses will be much greater concentration than the other. So basically what I'm trying to say is that at equilibrium the reactant and the products concentrations are almost equal. So that's going to be one of our answers for this problem. The second one is cackling for the concentration of N. 02. Let's first recognize that the equation for this equilibrium constant is equal to well it's a ratio. So for my numerator here it's the concentration of my products, brace the power of its coefficients over the concentration of my reactant, raise your power of its coefficients. So applying it to our chemical equation right over here We have the concentration of N 02 raised to the power of two Over the concentration of N 204. We're trying to isolate this right here because that's what we're solving for. So basically I'm going to do first is multiply both sides by our denominator. So we do So we get that the concentration of N. 02 raised to power of two is equal to our equilibrium constant multiplied by a concentration of our product or our star materials. So that's N. 204. They're finally isolating R. N. 02. Here, we're gonna go and scrub it both sides. So our concentration of N. Two and 02 is equal to, well R. K. C. Multiplied by the concentration of N 204. All raised to power of one over to order square rooted. So I'm gonna go ahead and actually plug in the numerical values for this. So he said that the equilibrium constant value is 0.212. Now for the concentration of N 204 were given this as well, that's 2.0 times 10 to the negative three molar. Again, this is all going to be square rooted. So once I put everything into a calculator, I see that my concentration of N. 02 is equal to 0.206 molar. So this right here is going to be my second and final answer for this problem.

Ch.15 - Chemical Equilibrium

All textbooks

Brown 15th Edition

Ch.15 - Chemical Equilibrium

Problem 14a

Chapter 15, Problem 14a

The equilibrium constant for the dissociation of molecular iodine, I₂(g) ⇌ 2 I(g), at 800 K is K_c = 3.1 × 10^–5. (a) Which species predominates at equilibrium I₂ or I?

Verified step by step guidance

Understand the equilibrium expression for the reaction: I2(g) ⇌ 2 I(g). The equilibrium constant, Kc, is given for this reaction.

Write the expression for the equilibrium constant Kc:

K_{c} = \frac{{[I]}^{2}}{[I2]}

. This expression relates the concentrations of the products and reactants at equilibrium.

Consider the value of Kc = 3.1 × 10^–5. A small Kc value indicates that the equilibrium position favors the reactants over the products.

Analyze the equilibrium position: Since Kc is much less than 1, the concentration of I2(g) is greater than the concentration of I(g) at equilibrium.

Conclude which species predominates: Given the small value of Kc, I2(g) predominates at equilibrium over I(g).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Equilibrium Constant (Kc)

The equilibrium constant, Kc, quantifies the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. For the dissociation of I2, Kc = [I]^2 / [I2]. A small Kc value, like 3.1 × 10^–5, indicates that at equilibrium, the concentration of reactants (I2) is much greater than that of the products (I), suggesting that the reaction favors the formation of reactants.

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. In the context of the dissociation of I2, if the concentration of I is increased or the temperature is altered, the equilibrium will shift to either the right (producing more I) or the left (producing more I2) to minimize the disturbance.

Predominance of Species at Equilibrium

To determine which species predominates at equilibrium, one must compare the concentrations of the reactants and products using the equilibrium constant. A Kc value significantly less than 1 indicates that the reactants are favored, meaning that at equilibrium, the concentration of I2 will be much higher than that of I, confirming that I2 predominates in this reaction.

Recommended video:

Guided course

03:03

Amphoteric Species

Related Practice

Textbook Question

The reaction A₂ + B₂ ⇌ 2 AB has an equilibrium constant K_c = 1.5. The following diagrams represent reaction mixtures containing A₂ molecules (red), B₂ molecules (blue), and AB molecules. (a) Which reaction mixture is at equilibrium?

views

Textbook Question

The diagram shown here represents the equilibrium state for the reaction A2(𝑔) + 2B(𝑔) ⇌ 2AB(𝑔). (a) Assuming the volume is 2 L, calculate the equilibrium constant 𝐾𝑐 for the reaction.

Textbook Question

Suppose that the gas-phase reactions A → B and B → A are both elementary reactions with rate constants of 4.7×10⁻³ s⁻¹ and 5.8×10⁻¹ s⁻¹, respectively. (a) What is the value of the equilibrium constant for the equilibrium A(g) ⇌ B(g)? (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

Textbook Question

The equilibrium constant for the dissociation of molecular iodine, I₂(g) ⇌ 2 I(g), at 800 K is K_c= 3.1×10⁻⁵. (b) Assuming both forward and reverse reactions are elementary reactions, which reaction has the larger rate constant, the forward or the reverse reaction?

views

Textbook Question

Write the expression for K_c for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous.

(a) 3 NO(g) ⇌ N₂O(g) + NO₂(g)

(b) CH₄(g) + 2 H₂S(g) ⇌ CS2(g) + 4 H₂(g)

(d) HF(aq) ⇌ H⁺(aq) + F^-(aq)

(e) 2Ag(s) + Zn²⁺(aq) ⇌ 2 Ag⁺(aq) + Zn(s)

(f) H₂O(l) ⇌ H⁺(aq) + OH^-(aq)

(g) 2 H₂O(l) ⇌ 2 H⁺(aq) + 2 OH^-(aq)

Textbook Question

Write the expressions for K_c for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous.

(a) 2 O₃(g) ⇌ 3 O₂(g)

(d) C(s) + 2 H₂(g) ⇌ CH₄(g)

(e) 4 HCl(aq) + O₂(g) ⇌ 2 H₂O(l) + 2 Cl₂(g)

(f) 2 C₈H₁₈(l) + 25 O₂(g) ⇌ 16 CO₂(g) + 18 H₂O(g)

(g) 2 C₈H₁₈(l) + 25 O₂(g) ⇌ 16 CO₂(g) + 18 H₂O(l)

The equilibrium constant for the dissociation of molecular iodine, I2(g) ⇌ 2 I(g), at 800 K is Kc = 3.1 × 10–5. (a) Which species predominates at equilibrium I2 or I?

Verified video answer for a similar problem:

Key Concepts

Equilibrium Constant (Kc)

Le Chatelier's Principle

Predominance of Species at Equilibrium

The equilibrium constant for the dissociation of molecular iodine, I₂(g) ⇌ 2 I(g), at 800 K is K_c = 3.1 × 10^–5. (a) Which species predominates at equilibrium I₂ or I?