Question

In the reaction $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$ the amount of each $PC{l}_5,PC{l}_3$ and $C{l}_2$ is 2 mole at equilibrium and total pressure is 3 atmosphere. The value of $k_p$ will be:

• A. 1.0 atm
• B. 3.0 atm
• C. 2.9 atm
• D. 6.0 atm

Answer 1

The correct option is A 1.0 atm

${PCl}_5\left(g\right)\rightleftharpoons {PCl}_3\left(g\right)+{Cl}_2\left(g\right)$ Total $P=3atm$

Given $2$ $2$ $2$

at Eq

$K_P$ of reaction $=\frac{P_{{Cl}_2}.P_{{PCl}_3}}{P_{{PCl}_5}}$

Partial pressure $=$ (mole fraction) $\times$ (total pressure)

${PCl}_2\Rightarrow \left(\frac{2}{6}\right)\times 3=1atm$

${PCl}_3\Rightarrow \left(\frac{2}{6}\right)\times 3=1atm$

${PCl}_5\Rightarrow \left(\frac{2}{6}\right)\times 3=1atm$

$K_P=\frac{1\times 1}{1}=1atm$