JEE MAIN - Chemistry (2022 - 24th June Evening Shift - No. 16)
PCl5 dissociates as
PCl5(g) $$\rightleftharpoons$$ PCl3(g) + Cl2(g)
5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is __________ $$\times$$ 10$$-$$3. (nearest integer)
(Given : R = 0.082 L atm K$$-$$1 mol$$-$$1; Assume ideal gas behaviour)
Explanation
_24th_June_Evening_Shift_en_16_1.png)
Here 2 moles of N2 also present that is why 2 moles always have to add in total mole calculation.
At equilibrium,
Pressure (P) = 2.46 atm
Volume (V) = 200 L
Temperature (T) = 600 K
$$\therefore$$ Applying ideal gas equation,
PV = nRT
$$\Rightarrow$$ 2.46 $$\times$$ 200 = (7 + x) $$\times$$ 0.082 $$\times$$ 600
$$\Rightarrow$$ x = 3
Now,
$${K_P} = {{{P_{PC{l_3}}} \times {P_{C{l_2}}}} \over {{P_{PC{l_5}}}}}$$
$$ = {{\left[ {{3 \over {7 + 3}} \times 2.46} \right]\left[ {{3 \over {7 + 3}} \times 2.46} \right]} \over {\left[ {{{5 - 3} \over {7 + 3}} \times 2.46} \right]}}$$
$$ = {{{3 \over {10}} \times {3 \over {10}} \times {{(2.46)}^2}} \over {{2 \over {10}} \times 2.46}}$$
$$ = {9 \over {20}} \times 2.46$$
$$ = 1107 \times {10^{ - 3}}$$ atm
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