Respuesta :
Answer:
Consider the following equilibrium:
2H2(g)+S2(g)⇌2H2S(g)Kc=1.08×107 at 700 ∘C.
What is Kp?
Explanation:
Given,
[tex]Kc=1.08 * 10^7[/tex]
The relation between Kp and Kc is:
[tex]Kp=Kc * (RT)^d^e^l^t^a^(^n^)[/tex]
Where delta n represents the change in the number of moles.
For the given equation,
The Delta n = Number of moles of products - number of moles of reactants
(2-(2+1))
=-1.
Hence,
Kp=Kc/RT.
Thus,
[tex]Kp=1.08 * 10^7 / 8.314 J.K6-1.mol^-^1 x 973 K\\Kp=1335.06[/tex]
The answer is Kp=1335.06
The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].
Explanation:
The relation between [tex]K_p \& K_c[/tex] is given by:
[tex]K_p=K_c(RT)^{\Delta n_g}[/tex]
Where:
[tex]K_c[/tex] = The equilibrium constant of reaction in terms of concentration
[tex]K_p[/tex] = The equilibrium constant of reaction in terms of partial pressure
R= The universal gas constant
T = The temperature of the equilibrium
[tex]n_g[/tex]= Change in gaseus moles
Given:
An equilibrium reaction, 700°C:
[tex]2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g),K_c=1.08\times 10^7[/tex]
To find:
The equilibrium constant in terms of partial pressure, [tex]K_p[/tex].
Solution:
The equilibrium constant of reaction in terms of concentration= [tex]K_c[/tex]
[tex]K_c=1.08\times 10^7[/tex]
The equilibrium constant of reaction in terms of partial pressure =[tex]K_p=?[/tex]
The gaseous moles of reactant side = [tex]n_r= 3[/tex]
The gaseous moles of product side = [tex]n_p= 2[/tex]
The temperature at which equilibrium is given = T
[tex]T = 700^oC+273.15 K=973.15K[/tex]
The change in gaseous mole = [tex]n_g=n_p-n_r=2-3 = -1[/tex]
[tex]K_p=1.08\times 10^7\times (0.0821 atm L/mol K\times 973.15 K)^{-1}\\K_p=1.35\times 10^5[/tex]
The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].
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