At 600° C , K p for the following reaction is 1 atm X(g) leftharpoons Y(g)+Z(g) at equilibrium, 50 % of X(g) is dissociated. The total pressure of the equilibrium system is P atm. What is the partial pressure (in atm) of X(g) at equilibrium?

Question

At 600° C , K p for the following reaction is 1 atm X(g) leftharpoons Y(g)+Z(g) at equilibrium, 50 % of X(g) is dissociated. The total pressure of the equilibrium system is P atm. What is the partial pressure (in atm) of X(g) at equilibrium? (A) 1 (B) 4 (C) 2 (D) 0.5

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/d73e0b884d56d0336917f81cc4286b12-.png" /> Kp=(Py ⋅ Pz/Px) 1=(( P /3) × ( P /3)/( P )3) ⇒ p =3 atm Partial pressure of X=(P/3)=1 atm

Q. At $60 0^{\circ} C, K_{p}$ for the following reaction is $1 a t m$
$X (g) ⇌ Y (g) + Z (g)$
at equilibrium, $50%$ of $X (g)$ is dissociated. The total pressure of the equilibrium system is $P$ atm. What is the partial pressure (in atm) of $X (g)$ at equilibrium?

1

4

2

0.5

$K_{p} = \frac{P _{y} \cdot P _{z}}{P _{x}}$

$1 = \frac{\frac{P}{3} \times \frac{P}{3}}{\frac{P}{3}}$

$\Rightarrow p = 3 a t m$

Partial pressure of $X = \frac{P}{3} = 1$ atm

Q. At 600∘C,Kp​ for the following reaction is 1atm X(g)⇌Y(g)+Z(g) at equilibrium, 50% of X(g) is dissociated. The total pressure of the equilibrium system is P atm. What is the partial pressure (in atm) of X(g) at equilibrium?

1

4

2

0.5

Kp​=Px​Py​⋅Pz​​ 1=3P​3P​×3P​​ ⇒p=3atm Partial pressure of X=3P​=1 atm

Q. At $60 0^{\circ} C, K_{p}$ for the following reaction is $1 a t m$
$X (g) ⇌ Y (g) + Z (g)$
at equilibrium, $50%$ of $X (g)$ is dissociated. The total pressure of the equilibrium system is $P$ atm. What is the partial pressure (in atm) of $X (g)$ at equilibrium?

$K_{p} = \frac{P _{y} \cdot P _{z}}{P _{x}}$

$1 = \frac{\frac{P}{3} \times \frac{P}{3}}{\frac{P}{3}}$

$\Rightarrow p = 3 a t m$

Partial pressure of $X = \frac{P}{3} = 1$ atm