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QJEEM22Chf608: JEE Main 2022 Chemistry - Class 11 JEEM22Chf608 Solution: 4.0 moles of argon and 5.0 moles of PCl 5 are introduced into…

4.0 moles of argon and 5.0 moles of PCl5 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 : R = 0.082 L atm K −- 1 mol −- 1]

Multiple Choice Options

  • A 2.25
  • B 6.24
  • C 12.13
  • D 15.24

Step-by-Step Verified Solution

JEE Main 2022 (Online) 29th June Evening Shift Chemistry - Chemical Equilibrium Question 40 English Explanation

Here 4 moles of inert gas argon also present.

∴\therefore Total moles of mixture present at equilibrium,

nT = 5 + x + 4

= 9 + x

At equilibrium, total pressure (pT) = 6 atm

Volume (v) = 100 L

Temperature = 610 K

∴\therefore Using ideal gas equation,

PTV=nTRT{P_T}V = {n_T}RT

⇒6×100=(9+x)×0.082×610\Rightarrow 6 \times 100 = (9 + x) \times 0.082 \times 610

⇒x=3\Rightarrow x = 3

Now,

KP=PPCl3×PCl2PPCl5{K_P} = {{{P_{PC{l_3}}} \times {P_{C{l_2}}}} \over {{P_{PC{l_5}}}}}

=[39+3×6]×[39+3×6][5−39+3×6]= {{\left[ {{3 \over {9 + 3}} \times 6} \right] \times \left[ {{3 \over {9 + 3}} \times 6} \right]} \over {\left[ {{{5 - 3} \over {9 + 3}} \times 6} \right]}}

=2712= {{27} \over {12}}

=94= {9 \over 4}

= 2.25 atm

Note :

Inert gas always contribute to total mole and pressure calculation.

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