AP EAMCET · Chemistry · Chemical Equilibrium
One mole of \(\mathrm{PCl}_5(\mathrm{~g})\) was heated in a \(1 \mathrm{~L}\) closed flask at \(500 \mathrm{~K}\). At equilibrium, 0.1 mole of \(\mathrm{Cl}_2(\mathrm{~g})\) was formed. What is its \(\mathrm{K}_{\mathrm{p}}\) (in atm)?
(Given \(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) )
- A \(2.7 \times 10^{-4}\)
- B 0.455
- C 0.0111
- D 90
Answer & Solution
Correct Answer
(B) 0.455
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Rightarrow \mathrm{K}_{\mathrm{c}}=\frac{\mathrm{x}^2}{1-\mathrm{x}}=\frac{(0.1)^2}{0.9}=0.011 \\ & \text { Thus, } \mathrm{K}_{\mathrm{p}}=(0.011)(0.082 \times 500)^{(2-1)} \\ & =0.451 .\end{aligned}\)
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