One mole of O_ 2(~g) and two moles of SO_ 2(~g) were heated in a closed vessel of one-litre capacity at 1098 ~K . At equilibrium 1.6 moles of SO_ 3(~g) were found. The equilibrium constant K_c of the reaction would be

Correct answer: C. 80 O_2+2 SO_2 leftharpoons 2 SO_3 at t=0 1 2 0 at equil. (1-α) 2(1-α) 2 α Given at equilibrium aligned & 2 α=1.6 & K_ C =((2 α)^2)/((1-α)(2-2 α)^2)= array c α=0.8 (0.8)^2 array (1-0.8)(1-0.8)^2 & =(0.64)/(0.002) & K_ C =80 & aligned

JEE Main 201226 MayOnline

Exam: JEE Main
Year: 2012
Date: 26 May
Mode: Online
Subject: Chemistry
Chapter: Chemical Equilibrium

JEE Main 2012 · 26 May · Online · Chemistry · Chemical Equilibrium

One mole of O_ 2(~g) and two moles of SO_ 2(~g) were heated in a...

One mole of

\mathrm{O}_{2(\mathrm{~g})}

and two moles of

\mathrm{SO}_{2(\mathrm{~g})}

were heated in a closed vessel of one-litre capacity at

1098 \mathrm{~K}

. At equilibrium

1.6

moles of

\mathrm{SO}_{3(\mathrm{~g})}

were found. The equilibrium constant

K_c

of the reaction would be

Options

Correct answer

C. 80

Step-by-step solution

\quad \mathrm{O}_2+2 \mathrm{SO}_2 \rightleftharpoons 2 \mathrm{SO}_3

at

\mathrm{t}=0 \quad 1 \quad 2 \quad 0

at equil.

(1-\alpha) 2(1-\alpha) \quad 2 \alpha

Given at equilibrium

\begin{aligned}& 2 \alpha=1.6 \\ & K_{\mathrm{C}}=\frac{(2 \alpha)^2}{(1-\alpha)(2-2 \alpha)^2}=\frac{\begin{array}{c} \alpha=0.8 \\ (0.8)^2 \end{array}}{(1-0.8)(1-0.8)^2}\\ & =\frac{0.64}{0.002}\\ & K_{\mathrm{C}}=80 \\ & \end{aligned}

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