JEE MAIN - Chemistry (2022 - 29th June Evening Shift - No. 3)
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]
Explanation
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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,
$${P_T}V = {n_T}RT$$
$$ \Rightarrow 6 \times 100 = (9 + x) \times 0.082 \times 610$$
$$ \Rightarrow x = 3$$
Now,
$${K_P} = {{{P_{PC{l_3}}} \times {P_{C{l_2}}}} \over {{P_{PC{l_5}}}}}$$
$$ = {{\left[ {{3 \over {9 + 3}} \times 6} \right] \times \left[ {{3 \over {9 + 3}} \times 6} \right]} \over {\left[ {{{5 - 3} \over {9 + 3}} \times 6} \right]}}$$
$$ = {{27} \over {12}}$$
$$ = {9 \over 4}$$
= 2.25 atm
Note :
Inert gas always contribute to total mole and pressure calculation.
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