Chemical equilibrium

Exercise 10

    

At a given temperature, the equilibrium constant for: $PCl_5(g)$ $PCl_3(g)$ $+$ $Cl_2(g)$ has the value of: $K_p$ $=$ $2,25$ atm Pure phosphorus pentachloride is introduced in a container with volume V and it appears that at equilibrium the partial pressure of this gas amounts to $0,25\;atm$ . - Calculate the partial pressures of the other two gases. -Calculate also the initial pressure of the phosphous pentachloride before dissociation.

$K_p$ $=$ $\frac{p_{PCl_3}p_{Cl_2}}{p_{PCl_5}}$ 1) Let $x$ be the partial pressure of $PCl_3$ at equilibrium. As each mole $PCl_5$ which is dissociated produces one mole $PCl_3$ and one mole $Cl_2$, these two gases must have the same number of moles at equilibrium, therefore $x$ is also the partial pressure of $Cl_2$ at equilibrium and so we have: $2,25$ $=$ $\frac{x\cdot x}{0,25} $ $x$ $=$ $\sqrt{2,25\cdot 0,25}$ $=$ $0,75$ atm At equilibrium: $p_{PCl_3}$ $=$ $p_{Cl_2}= 0,75$ atm 2) At constant pressure and temperature, the partial pressures are proportional to the numbers of moles. Therefore partial pressures can be treated exactly as the number of moles: If $0,75$ atm of $PCl_3$ have appeared, then $0,75$ atm of $PCl_5$ must have disappeared. The initial pressure of $PCl_5$ was therefore equal to $0,25$ (the actual pressure) + $0,75= 1 $ atm