Question

The equilibrium constant for the decomposition of PCl5 at 250 celcius is 1.05. PCl5(g)-->PCl3(g)+Cl2(g) If the equilibrium pressures of PCl5 and PCl3 are .875 atm and .463 atm, respectively, then what is the equilibrium partial pressure of Cl2 at 250 celsius.

Answer 1

Answer : The equilibrium partial pressure of Cl₂ at 25°C is, 1.98 atm

Explanation :

For the given chemical reaction:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{P_{PCl_3}\times P_{Cl_2}}{P_{PCl_5}}$

We are given:

$P_{PCl_5}=0.875atm$

$P_{PCl_3}=0.463atm$

$K_p=1.05$

Putting values in above equation, we get:

$1.05=\frac{0.463\times P_{Cl_2}}{0.875}\\\\P_{Cl_2}=1.98atm$

Thus, the equilibrium partial pressure of Cl₂ at 25°C is, 1.98 atm