The equilibrium constant, kp, for the following reaction is 9.52ã10-2 at 350 k: ch4(g) + ccl4(g) 2ch2cl2(g). calculate the equilibrium partial pressures of all species when ch4 and ccl4, each at an intitial partial pressure of 0.844 atm, are introduced into an evacuated vessel at 350 k. a) pch4 = atm. b) pccl4 = atm. c) pch2cl2 = atm.

A) p CH₄       = 0.732 atm

B) p CCl₄      = 0.732 atm

c)  p CH₂Cl₂  = 0.22 atm

Explanation:

we have the equilibrium constant for this problem, along the initial pressures for the reactants, and  need to find the partial pressures at equilibrium. So lets setup the equilibrium:

CH₄ (g) + CCl₄ (g)   ⇔  2 CH₂Cl₂ (g)  Kp =  9.50 x 10⁻²

where Kp is given by :

Kp = p CH₂Cl₂ ² / p CH₄ x p CCl₄

where p are the partial pressures

                               p CH₄  atm          p CCl₄  atm              p CH₂Cl₂  atm

initial                             0.844              0.844                           0

change                           - x                     - x                           +2x

equilibrium               0.844 - x            0.844 - x                      2 x

Kp = 9.52 x 10⁻² = ( 2x )²/  (( 0.844 - x ) x ( 0.844 - x ))

9.52 x 10⁻² =  (2x)² / ( 0.844 - x )²

Taking square root to both sides of the equation:

√9.52 x 10⁻²  = 2x / (0.844 - x )

0.309 = 2x / (0.844 - x)

0.260 - 0.309 x = 2x

0.260 = 2.309 x   ⇒ x = 0.112

So the partial pressures are at equilibrium are:

p CH₄ = p CCl₄ = 0.844 - 0.112 = 0.732 atm

p CH₂Cl₂  = 2 x (0.112 atm) = 0.224 atm

You can check your answer is correct by plugging this values and comparing  with the given Kp.

no but i do not usually comment or use my voice that much when it comes to social media