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QJEEM22Ch8a3f: JEE Main 2022 Chemistry - Class 11 JEEM22Ch8a3f Solution: PCl 5 dissociates as PCl 5 (g) PCl 3 (g) + Cl 2 (g) 5 moles of…

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)

Step-by-Step Verified Solution

JEE Main 2022 (Online) 24th June Evening Shift Chemistry - Chemical Equilibrium Question 35 English Explanation

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,

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

=[37+3×2.46][37+3×2.46][5−37+3×2.46]= {{\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]}}

=310×310×(2.46)2210×2.46= {{{3 \over {10}} \times {3 \over {10}} \times {{(2.46)}^2}} \over {{2 \over {10}} \times 2.46}}

=920×2.46= {9 \over {20}} \times 2.46

=1107×10−3= 1107 \times {10^{ - 3}} atm

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