For the reaction: PCl₃(g)+ Cl₂(g)⇌ PCl₅(g)at 70.5 °C,K_p = 1.05. If one starts with 1.80 atm pressure of PCl₃(g),1.72 atm pressure of Cl₂(g),and no PCl₅(g),what is the partial pressure of PCl₅(g)at equilibrium?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is $K_p = \frac{[PCl_5]}{[PCl_3][Cl_2]}$. Given $K_p = 1.05$, initial pressures of $PCl_3 = 1.80$ atm and $Cl_2 = 1.72$ atm, and $PCl_5 = 0$ atm, we can set up an ICE table. At equilibrium, the pressures change by $x$: $PCl_3$ and $Cl_2$ decrease by $x$, and $PCl_5$ increases by $x$. Thus, $K_p = \frac{x}{(1.80-x)(1.72-x)} = 1.05$. Solving this equation for $x$ gives the change in concentration, and since $PCl_5$ starts at 0 atm, its equilibrium pressure is equal to $x$, which is approximately 0.856 atm.

For the Reaction: PCl3(g)+ Cl2(g)⇌ PCl5(g)at 70.5 °C,Kp = 1.05

For the Reaction: PCl₃(g)+ Cl₂(g)⇌ PCl₅(g)at 70.5 °C,K_p = 1.05