JEE MAIN - Chemistry (2021 - 27th July Morning Shift - No. 21)
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)
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)
Answer
1400
Explanation
PCl5(g) $$\rightleftharpoons$$ PCl3(g) + Cl2(g) K2 = 1.844
t = 0 3moles
t = $$\infty$$ x x
$$ \Rightarrow {{[PC{l_3}][C{l_2}]} \over {[PC{l_5}]}} = {{{x^2}} \over {3 - x}} = 1.844$$
$$ \Rightarrow {x^2} + 1.844 - 5.532 = 0$$
$$ \Rightarrow x = {{ - 1.844 + \sqrt {{{(1.844)}^2} + 4 \times 5.532} } \over 2}$$
$$ \cong 1.604$$
$$\Rightarrow$$ Moles of PCl5 = 3 $$-$$ 1.604 $$\cong$$ 1.396
t = 0 3moles
t = $$\infty$$ x x
$$ \Rightarrow {{[PC{l_3}][C{l_2}]} \over {[PC{l_5}]}} = {{{x^2}} \over {3 - x}} = 1.844$$
$$ \Rightarrow {x^2} + 1.844 - 5.532 = 0$$
$$ \Rightarrow x = {{ - 1.844 + \sqrt {{{(1.844)}^2} + 4 \times 5.532} } \over 2}$$
$$ \cong 1.604$$
$$\Rightarrow$$ Moles of PCl5 = 3 $$-$$ 1.604 $$\cong$$ 1.396
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