If K c = 7.5×10 −9 at 1000 K for the reaction N 2 (g) + O 2 (g) → 2 NO(g), give the value of K c at 1000 K for the reaction (b) NO(g) → 1/2 N 2 (g) + 1/2 O 2 (g)

Welcome back everyone. At 298 Kelvin, the equilibrium constant for the reaction, two moles of carbon monoxide plus one mole of oxygen produces two moles of carbon dioxide is two multiplied by 10. The power of 11th determine the equilibrium constant for the reaction. Two moles of co two in equilibrium with two moles of CO O plus 02 at 298 Kelvin. So let's analyze the two equilibria reactions given to us. The first one given to us is two moles of carbon monoxide plus one mole of oxygen produce two moles of carbon dioxide. Let's call it K one. We know that it is given and our target equation is two moles of carbon dioxide in equilibrium with two moles of carbon monoxide and one mole of oxygen. Let's call the equilibrium constant for this reaction. K two. What we notice is that reaction number two is actually the flipped version of reaction. Number one, it's basically reverse right? Because our reactants from reaction number one now become our products and our products from reaction number one become our reactants. So what is the relationship between K one and K two. Well, essentially, whenever we are reversing a reaction, we want to take the reciprocal value of the equilibrium constant. So K two is simply equal to one divided by K one, right. And therefore we have our expression. So the equilibrium constant of interest K two is simply one divided by two, multiplied by 10, the power of 11th, which gives us five multiplied by sense, the power of negative T 12. And that's our final answer for the equilibrium constant B Thank you for watching.

Ch.15 - Chemical Equilibrium

McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook

All textbooks

McMurry 8th Edition

Ch.15 - Chemical Equilibrium

Problem 15.60b

Chapter 15, Problem 15.60b

If K_c = 7.5×10⁻⁹ at 1000 K for the reaction N₂(g) + O₂(g) → 2 NO(g), give the value of K_c at 1000 K for the reaction
(b) NO(g) → 1/2 N₂(g) + 1/2 O₂(g)

Verified step by step guidance

Identify the given equilibrium constant \( K_c = 7.5 \times 10^{-9} \) for the reaction \( \text{N}_2(g) + \text{O}_2(g) \rightarrow 2 \text{NO}(g) \).

Recognize that the reaction in part (b) is the reverse of the given reaction, but with coefficients halved: \( \text{NO}(g) \rightarrow \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \).

Recall that when a reaction is reversed, the equilibrium constant for the reverse reaction is the reciprocal of the original reaction's equilibrium constant.

Understand that when the coefficients of a balanced equation are multiplied by a factor, the equilibrium constant is raised to the power of that factor.

Calculate the new equilibrium constant by taking the reciprocal of \( K_c \) and then raising it to the power of \( \frac{1}{2} \) to account for the change in coefficients.

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) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. It is calculated using the formula Kc = [products]^[coefficients] / [reactants]^[coefficients]. A small Kc value, like 7.5×10^−9, indicates that at equilibrium, the reaction favors the reactants over the products.

Reaction Quotient (Q)

The reaction quotient (Q) is a measure of the relative concentrations of products and reactants at any point in time during a reaction. It is calculated in the same way as Kc but does not require the system to be at equilibrium. Comparing Q to Kc helps predict the direction in which a reaction will proceed to reach equilibrium.

Reverse Reaction and Kc Relationship

For a given reaction, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. If Kc for the forward reaction is known, Kc for the reverse reaction can be calculated as Kc(reverse) = 1/Kc(forward). This relationship is crucial for determining Kc for the reaction given in the question, which is the reverse of the original reaction.

Recommended video:

Guided course