Answer:
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.
