Question

For the reaction PCl5(g) ⇔ PCl3(g) + Cl2(g), the moles of each component PCl5, PCl3 and Cl2 at equilibrium were found to be 2. If the total pressure is 3 atm. The Kp will be​

Answer 1

Answer:

1 atm.

Explanation:

pcl5 = pcl3 + cl2

2.0 2.0 2.0

than partial pressure of pcl5 = 2/6 ×3

partial pressure of pcl3 = 2/6×3

partial pressure of cl2 = 2/6×3

than kp = 1×1÷1 =1

Answer 2

Answer:

The equilibrium constant fro the reaction is 1.

Explanation:

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

Moles of $PCl_5$ at equilibrium = 2 mol

Moles of $PCl_3$ at equilibrium = 2 mol

Moles of $Cl_2$ at equilibrium = 2 mol

Mole fraction of $PCl_5$ :

$\chi_1=\frac{2 mol}{2 mol +2mol+2 mol}=\frac{1}{3}$

Mole fraction of $PCl_3$ :

$\chi_2=\frac{2 mol}{2 mol +2mol+2 mol}=\frac{1}{3}$

Mole fraction of $Cl_2$ :

$\chi_3=\frac{2 mol}{2 mol +2mol+2 mol}=\frac{1}{3}$

Partial pressure of $PCl_5$ at equilibrium =$p_1$

Partial pressure of$PCl_3$ at equilibrium = $p_2$

Partial pressure of $Cl_2$ at equilibrium = $p_3$

Total pressure at equilibrium = 3 atm

$p_1=p\times \chi_1=3 atm\times \frac{1}{3}=1 atm$

$p_2=p\times \chi_2=3 atm\times \frac{1}{3}=1 atm$

$p_3=p\times \chi_3=3 atm\times \frac{1}{3}=1 atm$

The equilibrium constant will be given as:

$K_p=\frac{p_2\times p_3}{p_1}=\frac{1 atm \times 1 atm}{1 atm}= 1$