Respuesta :
Answer:
Kc = 2300 L·Atm·mol⁻¹
Explanation:
4HCl(g) + O₂(g) => 2Cl₂(g) + 2H₂O(g)
P(i) 2.3 atm 1.00 atm 0 0
ΔP -4x -x +2x +2x
P(eq) 2.3-4x 1.00-x 2x 2x
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Given P(Cl₂) at equil. = 0.93 atm => 2x = 0.93 => x = 0.465 atm
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P(eq) [2.3-4(0.465)]atm 1.00-0.465atm 2(0.465)atm 2(0.465)atm
= 0.44 atm = 0.535 atm = 0.93 atm =0.93 atm
Kp = (P(Cl₂))²(P(H₂O))²/(P(HCl))⁴(P(O₂))¹ = (0.93)²(0.93)²/(0.44)⁴(0.535)¹ atm⁻¹ = 37.3 atm⁻¹
Kc = Kp(R·T)^-Δn = 37.3(0.08206)(750)⁻⁽⁻¹⁾ L·atm·mol⁻¹ = 2296 L·atm·mol⁻¹ ≅ 2300 L·atm·mol⁻¹
The Kp of the reaction is 37.3
data;
- Pressure of HCL = 2.30atm
- Pressure of O2 = 1.00atm
- Temperature = 750K
- Equilibrium pressure of Cl2 = 0.93 atm
Equilibrium Pressure
From the equation of reaction
[tex]4HCL + O_2 \to 2Cl_2 +2H_2O[/tex]
Initial 2.3 1.0 0 0
change -4x -1x +2x +2x
equilibrium 2.3-4x 1.0 - 1x +2x +2x
At equilibrium, the pressure of chlorine is given as 0.93atm
Equilibrium constant expression is given as
[tex]K_p = \frac{[Cl]^2[H_2O]^2}{[HCl]^4[O_2]} \\K_p = \frac{[2x]^2 [2x]^2}{[2.3-4x]^4[1.0-1x]}[/tex]
solving the above equation, the value of Kp is 37.3
The Kp of the reaction is 37.3
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