4.0 moles of argon ...|CHEMICAL EQUILIBRIUM|JEE-Main 2022

JEE MAIN 2022

29-06-2022 S2

Question

4.0 moles of argon and 5.0 moles of $\mathrm{PCl}_5$ are introduced into an evacuated flask of 100 litre capacity at 610 K . The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm . The Kp for the reaction is [Given: $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]

Select the correct option:

2.25

6.24

12.13

15.24

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & \mathrm{PCl}_5=5 \mathrm{~mole} \\ & \mathrm{Ar}=4 \mathrm{~mole} \\ & \mathrm{P}_{\text {Total }}=\frac{9 \times 0.82 \times 610}{100}=4.5 \mathrm{~atm} \\ & \mathrm{P}_{\mathrm{PCl}_5}=\frac{5 \times 4.5}{9}=2.5 ; \mathrm{P}_{\mathrm{Ar}}=\frac{4 \times 4.5}{9}=21 \\ & \mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2 \\ & 2.5-\mathrm{P} \quad \mathrm{P} \quad \mathrm{P} \\ & \mathrm{P}_{\text {total }}=2.5-\mathrm{P}+\mathrm{P}+\mathrm{P}+\mathrm{P}_{\text {AAL }}=6 \\ & \mathrm{P}=1.5\end{aligned}$
$K_p=\frac{1.5 \times 1.5}{1}=2.25$

Question Tags

JEE Main

Chemistry

Medium

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