Verified Solution • 2025–26 Session moderate

QJEEM21Ch2222: JEE Main 2021 Chemistry - Class 11 JEEM21Ch2222 Solution: PCl 5 PCl 3 + Cl 2 K c = 1.844 3.0 moles of PCl 5 is introduced …

PCl5 ⇌\rightleftharpoons PCl3 + Cl2

Kc = 1.844

3.0 moles of PCl5 is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of PCl5 at equilibrium is ______________ ×\times 10 −- 3. (Round off to the Nearest Integer)

Step-by-Step Verified Solution

PCl5(g) ⇌\rightleftharpoons PCl3(g) + Cl2(g) K2 = 1.844

t = 0 3moles

t = ∞\infty x x

⇒[PCl3][Cl2][PCl5]=x23−x=1.844\Rightarrow {{[PC{l_3}][C{l_2}]} \over {[PC{l_5}]}} = {{{x^2}} \over {3 - x}} = 1.844

⇒x2+1.844−5.532=0\Rightarrow {x^2} + 1.844 - 5.532 = 0

⇒x=−1.844+(1.844)2+4×5.5322\Rightarrow x = {{ - 1.844 + \sqrt {{{(1.844)}^2} + 4 \times 5.532} } \over 2}

≅1.604\cong 1.604

⇒\Rightarrow Moles of PCl5 = 3 −- 1.604 ≅\cong 1.396

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