Respuesta :
Kp = 0.77 in the equilibria H₂(g) + CO₂(g) ⇄ H₂O(g) + CO(g); Kc = 0.77 at 1020K
Explanation
We know that Kp = [tex]Kc(RT)^{\Delta n}[/tex]
Lets find Δn
Δn = [tex]n_{gas}[/tex] on the product side - [tex]n_{gas}[/tex] on the reactants side
= (1 + 1) - (1 + 1)
= 0
So, now we have
Kc = 0.77
R = 0.08206 L.atm/K.mol
T = 1020K
Δn = 0
So Substituting the value in Kp = [tex]Kc(RT)^{\Delta n}[/tex] we get
Kp = 0.77(0.08206 × 1020K)⁰
Kp = 0.77(1)
Kp = 0.77
What is Kp?
Kp is an equilibrium constant depending on partial pressure. The ratio of reactants to products in a reaction at equilibrium is disclosed.
A few Kp characteristics are:
- Kp is a partial pressure-based equilibrium constant. It provides information on the ratio of reactants to products in a reaction at equilibrium.
- While the value of Kp is unaffected by changes in concentration or pressure, the equilibrium does shift as a result.
- The position and Kp value of a gaseous equilibrium change when the equilibrium's temperature shifts.
- The presence of a catalyst has no impact on the value of Kp or the location of the equilibrium.
- Kp and Kc are very comparable. Kc uses molar concentrations while Kp here uses partial pressures.
Learn more about Kc
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