Consider the reaction where Kp = 0.497 at 500 K. PCl5(g) leftharpoons PCl3(g) +Cl2(g) If the three gases are mixed in a rigid container so that the partial pressure of each gas is initially 1 atm. Which is true?

Question

Consider the reaction where Kp = 0.497 at 500 K. PCl5(g) leftharpoons PCl3(g) +Cl2(g) If the three gases are mixed in a rigid container so that the partial pressure of each gas is initially 1 atm. Which is true? (A) More PCl5 will be produced. (B) More PCl5 will not be produced. (C) Equilibrium will be established when 50% reaction is complete (D) None of the above

Tardigrade Admin · Accepted Answer

Qp = (PPCl3 × PCl2 /PPCl5) =(1 atm × 1 atm/1 atm)=1

Q. Consider the reaction where $K_{p} = 0.497$ at 500 K.
$PC l_{5} (g) ⇌ PC l_{3} (g) + C l_{2} (g)$
If the three gases are mixed in a rigid container so that the partial pressure of each gas is initially 1 atm. Which is true?

More $PC l_{5}$ will be produced.

More $PC l_{5}$ will not be produced.

Equilibrium will be established when 50% reaction is complete

None of the above

$Q_{p} = \frac{P _{PC l_{3}} \times P _{C l_{2}}}{P _{PC l_{5}}}$
$= \frac{1 a t m \times 1 a t m}{1 a t m} = 1$

Q. Consider the reaction where Kp​=0.497 at 500 K. PCl5​(g)⇌PCl3​(g)+Cl2​(g) If the three gases are mixed in a rigid container so that the partial pressure of each gas is initially 1 atm. Which is true?

More PCl5​ will be produced.

More PCl5​ will not be produced.