Calculating equilibrium constant Kp using partial pressures (article) | Khan Academy (2024)

Definition of equilibrium constant Kp for gas phase reactions, and how to calculate Kp from Kc.

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  • gizem.cetin.usa

    8 years agoPosted 8 years ago. Direct link to gizem.cetin.usa's post “Why do we not include liq...”

    Why do we not include liquids and solids in equilibrium expressions? What do you mean by "their concentration is equal to 1"?

    (19 votes)

    • Alex Wrede

      8 years agoPosted 8 years ago. Direct link to Alex Wrede's post “The concentration of the ...”

      Calculating equilibrium constant Kp using partial pressures (article) | Khan Academy (4)

      Calculating equilibrium constant Kp using partial pressures (article) | Khan Academy (5)

      The concentration of the solids never changes, as its density remains the same. Regarding liquids, adding or removing liquids has an insignificant effect on the concentration of the system, as the system is in an aqueous solution. Yes, the liquids do change the concentration, but not by a measurable amount. It's like adding another fish to the sea, making pretty much no difference to the concentration of fish in the ocean

      (35 votes)

  • Mushahid Osman

    8 years agoPosted 8 years ago. Direct link to Mushahid Osman's post “In example 2, how do you ...”

    In example 2, how do you know that you are finding the constant in terms of bar and not atm?

    (7 votes)

  • caroxxwu

    6 years agoPosted 6 years ago. Direct link to caroxxwu's post “For the equation Kp = Kc(...”

    For the equation Kp = Kc(RT)^(delta N), shouldn't there be two instances in which Kp = Kc? First, when delta N = 0 (mols of product gas = mols of reactant gas); second when temperature T is the exact reciprocal of constant R or when R*T = 1 (if R = 0.08206 L*atm*mol^(-1)*K^(-1), T = 1/0.08206 K)? What is the significance of Kp = Kc?

    (7 votes)

  • Clara I-W

    8 years agoPosted 8 years ago. Direct link to Clara I-W's post “Can anyone explain why Pr...”

    Can anyone explain why Pressure does not change the equilibrium constant for a reaction but temperature does?

    (3 votes)

    • Ernest Zinck

      8 years agoPosted 8 years ago. Direct link to Ernest Zinck's post “The equilibrium constant ...”

      The equilibrium constant is a constant.
      If you change a pressure, all the pressures change so that their new values will give the same value for the equilibrium constant.

      (7 votes)

  • AJ

    7 years agoPosted 7 years ago. Direct link to AJ's post “For the ICE Table, how do...”

    For the ICE Table, how do we know that the Change has +2x and +x? I mean, why did we choose plus instead of minus?

    (4 votes)

    • RogerP

      7 years agoPosted 7 years ago. Direct link to RogerP's post “With the ICE table, you i...”

      With the ICE table, you initially assume that the reaction is entirely on the left, with water but no hydrogen or oxygen gases. Then, as the reaction proceeds, hydrogen and oxygen are formed, which are written as +2x and +x atm in the "change" line.

      If you used minus signs, that would indicate that the gas concentrations were decreasing as the reaction moved towards equilibrium, which is not the case.

  • Adit Khokar

    8 years agoPosted 8 years ago. Direct link to Adit Khokar's post “in the formula Kc = Kp * ...”

    in the formula Kc = Kp * (RT)^n
    if we use double the reactants then the difference b/w the number of moles will also get double Kc will change

    (2 votes)

    • ŞΣΔşђ ĐŗДġΦЇđ

      8 years agoPosted 8 years ago. Direct link to ŞΣΔşђ ĐŗДġΦЇđ's post “but you cant arbitrarily ...”

      but you cant arbitrarily double the moles of reactants or products in a particular BALANCED reaction.
      all values for the moles should be calculated using a balanced equation simplified to its lowest form.
      increasing quantity of reactants will not change the value of kc either, kc always is a constant for a given temperature.
      but it has no effect on difference in moles for a balanced reaction.
      (correct me if i'm wrong)

      (5 votes)

  • Lucian Rex

    7 years agoPosted 7 years ago. Direct link to Lucian Rex's post “When the do the PV=nRT, s...”

    When the do the PV=nRT, shouldn't the final equation be P=nRT/V, not P=(n/V)RT?

    (1 vote)

    • yuki

      7 years agoPosted 7 years ago. Direct link to yuki's post “The two expressions P=nRT...”

      The two expressions P=nRT/V and P=(n/V)RT are the same. The second one is written specifically to separate out the expression n/V, which is equal to the concentration. This helps emphasize the relationship between concentration and the ideal gas equation.

      (5 votes)

  • apndolby297

    8 years agoPosted 8 years ago. Direct link to apndolby297's post “As in example 3, why are ...”

    As in example 3, why are we taking the partial pressure of water in liquid form as 1, why not any other value (say less than 1)?

    (2 votes)

    • ŞΣΔşђ ĐŗДġΦЇđ

      8 years agoPosted 8 years ago. Direct link to ŞΣΔşђ ĐŗДġΦЇđ's post “adding or removing liquid...”

      adding or removing liquid makes a very negligible difference in the pressure of the system.
      you can assume it as 1 without changing the answer.
      at least for calculation purposes
      besides taking any value apart from 1 will unnecessarily increase the difficulty of your calculations.

      (3 votes)

  • Sneha Shrestha

    7 years agoPosted 7 years ago. Direct link to Sneha Shrestha's post “why it is difficult to fi...”

    why it is difficult to find equilibrium constant at room temperature

    (3 votes)

    • Anshari Hasanbasri

      6 years agoPosted 6 years ago. Direct link to Anshari Hasanbasri's post “Why is it difficult? You ...”

      Why is it difficult? You just need to plug in the room temperature for T in Kp=Kc(RT)^(delta n) (assuming you were to find Kp from Kc of the molecules)

      (1 vote)

  • Swarna

    7 years agoPosted 7 years ago. Direct link to Swarna's post “In example 3, what does N...”

    In example 3, what does N/A in the ICE table indicate?

    (2 votes)

    • RogerP

      7 years agoPosted 7 years ago. Direct link to RogerP's post “N/A is a common abbreviat...”

      N/A is a common abbreviation for "Not applicable".

      (2 votes)

Calculating equilibrium constant Kp using partial pressures (article) | Khan Academy (2024)

FAQs

How do you calculate the equilibrium constant K when given the equilibrium partial pressures? ›

For the equation a b c d A + B C + D , 𝐾  is calculated as the product of the partial pressures C and D divided by the product of the partial pressures of A and B , where each individual partial pressure is raised to the power of their respective stoichiometric coefficients c, d, a, and b.

How to write the equilibrium constant expression in terms of partial pressure? ›

Kp is defined as Πj(PjPΘ)νj where Pj is the partial pressure of component j and ν is the stoichiometric coefficient. I have a couple of questions already. Firstly, Kp is meant to be dimensionless so therefore, PjPΘ must be dimensionless.

What is the equilibrium constant with respect to partial pressure KP? ›

Kp is an equilibrium constant based on partial pressures. It tells you the ratio of products to reactants in a reaction at equilibrium. Partial pressure is the pressure that a single constituent gas exerts on a closed system.

What is the formula for KP KC equilibrium? ›

Generally, the relation between Kp and Kc can be represented as: Kp = Kc (RT) Kc = Kp(RT)-

How to calculate KP from Gibbs free energy? ›

The standard Gibbs free energy change, ΔG0 is related to equilibrium constant, Kp as: Kp=−RTlnΔG0. Kp=[eRT]ΔG0. Kp=e−ΔG0/RT.

What is K in terms of partial pressures? ›

\text{K}_p is a value that shows the relationship between the partial pressure of the reactants and the partial pressure of the products. \text{K}_p is calculated by dividing the partial pressures of the products by the partial pressures of the reactants.

How to write kp expression? ›

Writing an expression for Kp

Kp has exactly the same format as Kc, except that partial pressures are used instead of concentrations. The gases on the right-hand side of the chemical equation are at the top of the expression, and those on the left at the bottom.

What is the relationship between the equilibrium constant and the equilibrium pressure? ›

The equilibrium constant is a constant. If you change a pressure, all the pressures change so that their new values will give the same value for the equilibrium constant.

What is the equilibrium constant unit for KP? ›

Therefore, the unit for Kp= atm or bar.

How to calculate equilibrium constant? ›

Just dividing the product concentrations or pressures at their stoichiometric coefficients by the reactant concentrations or pressures at their stoichiometric coefficients gives you the equilibrium constant.

What does the equilibrium constant KP depend on? ›

Kp is affected by changes in temperature, which can shift the position of the equilibrium and change the value of Kp. Increasing the temperature favors the endothermic reaction and increases the partial pressure of the products, leading to an increase in Kp.

How do you find K when given the equilibrium constant? ›

Write the equilibrium constant expression for the reaction. Substitute the known K value and the final concentrations to solve for x. Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain K.

What is the K in the equilibrium equation? ›

The ratio of the rate constants for the forward and reverse reactions at equilibrium is the equilibrium constant (K), a unitless quantity. The composition of the equilibrium mixture is therefore determined by the magnitudes of the forward and reverse rate constants at equilibrium.

How do you calculate the thermodynamic equilibrium constant K? ›

The relation K=exp(−ΔrG∘/RT) (Eq. 11.8. 11) gives us a way to evaluate the thermodynamic equilibrium constant K of a reaction at a given temperature from the value of the standard molar reaction Gibbs energy ΔrG∘ at that temperature. If we know the value of ΔrG∘, we can calculate the value of K.

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