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At high temperature, HCl and O2 react as 4HCl(g) O2(g) 2Cl2(g) 2H2O(g) If HCl at 2. 30 atm and O2(g) at 1. 00 atm react at 75 0 K, the equilibrium pressure of Cl2(g) is 0. 93 atm. Calculate K.

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drpelezo

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

Learn more on equilibrium pressure kp here;

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