Question

for the reaction pcl3 + cl2=pcl5 at 250°C, the value of Kc=26, then the value of Kp at the same temperature will be?

Answer 1

Answer: The value of $K_p$ at 250°C is 0.606

Explanation:

The give chemical reaction follows:

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

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = 26

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $250^oC=250+273=523K$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=1-2=-1$

Putting values in above equation, we get:

$K_p=26\times (0.0821\times 523)^{-1}\\\\K_p=0.606$

Hence, the value of $K_p$ at 250°C is 0.606

Answer 2

Answer:

your answer is in the above attachment