What does it mean if Kp is less than or greater than 1?
Challenge Question => When does Kp = Kc?
So I will join in. First, I think the first response is OK and what follows, in my opinion, is much ado about not much.To clarify my short response, I just wanted to point out, and for no particular reason, that the question was about Kp but the answer was about Kc. The reasoning, however, is the same, If Kc>1 or Kp>1 then product side exceed reactant side. As for when Kp = Kc, if delta moles is 0, then Kp = Kc(RT)^dn or Kp = Kc(RT)^0 or Kp = Kc(1) or Kp = Kc.
I'm outta here.
Right on! Dr Bob:-)
OK … When Σ molar volumes reactants = Σ molar volumes products => Kp = Kc
Standard equation relating Kp and Kc is Kp = Kc(RT)^Δn where Δn = change in total molar volumes of reaction = Vₘ(Products) - Vₘ(Reactants).
3H₂(g) + N₂(g) => 2NH₃(g); Δn = 2Vₘ - 4Vₘ = -2Vₘ => Kp ≠ Kc
H₂(g) + Cl₂(g) => 2HCl(g) => Δn = 0 => Kp = Kc(RT)ᵒ = Kc(1) => Kp = Kc
I would concede to your comment if the responses to the original question were correct, but they are NOT.
The equilibrium constant is a numerical value of a reaction that defines its 'Extent of Reaction'; that is, how long it takes for the reactants to go through the kinetic startup and achieve the steady state dynamic equilibrium where the rate of the forward reaction equals the rate of the reverse reaction. Equating the rate law expressions for the forward and reverse reactions and solving for rate constant ratios that express increasing values with increasing product yield is chosen as the appropriate expression for calculating the equilibrium constant. This is published as follows:
The reaction equilibrium constant is the product of the concentrations of products raised to their respective coefficient values of the balanced standard equation divided by the product of the concentrations of the reactants raised to their respective coefficient values of the balanced standard equation.
This gives an expression that relates to the reaction yield and not simply the amount of products vs reactants.
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Example: Consider the reaction 2SO₂(g) + O(g) => 2SO₃(g); Kc = 1558 at 900K => In this reaction the amount of products (=> 2 Molar Volumes SO₃(g)) is LESS THAN the amount of reactants (=> 3 Molar Volumes SO₂(g) + 1 Molar Volume O₂(g)) yet the Kc-value is significantly above 1 and contradicts the reply ‘When Kc is greater than 1, products exceed reactants (at equilibrium).
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Also, for sake of interest, here’s a problem that depends on one having a basic knowledge of the relationship between Kp and Kc. Similar problems are found in both high school and college level text books. Here’s the Kp / Kc problem…
A 3.00-mol sample of HI(g) is placed into a 1.00-L vessel at 460ᵒ, and the reaction system is allowed to come to equilibrium according to the reaction 2HI(g) => H₂(g) + I₂(g). The HI(g) partially decomposes, forming 0.229-mol H₂ and 0.229 mol I₂ at equilibrium. What is Kp for the following reaction at 460ᵒ?
½H₂(g) + ½I₂(g) => HI(g)
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My comment was given as an open contribution to all who might be studying chemical equilibrium. Use it if you wish or not. It was not a complex dissertation on Chem Equilibrium. AND, for future reference, please avoid being condescending in your replies, it really shows a lack of class and does nothing for the collaborative work on this site. To be blunt, there’s no difference between high school general chemistry and college level general chemistry. Both are traditionally taught from the Dalton perspective and evolve from the simple to the more complex. Both contain strong presentations of Chemical Kinetics, Equilibrium & Thermodynamic study of reactions. The only difference is one (high school chemistry) comes before college chemistry in curricula. Oh, and no one attempted to offer a response to the question when is Kp = Kc. … So, I challenge you to work the HI(g) decomposition reaction problem above, and present your work. It might expand your horizons. 😊