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JEE Main 202120 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
20 July
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2021 · 20 July · Morning Shift · Chemistry · Chemical Equilibrium

2 SO _ 2 ( g ) + O _ 2 ( g ) ⇌ 2 SO _ 3 ( g ) In an equilibrium...

2SO2(g)+O2(g)⇌2SO3(g)2{SO}_{2} ( g ) + {O}_{2} ( g ) ⇌ 2{SO}_{3} ( g )
In an equilibrium mixture, the partial pressures are
PSO3=43kPa;PO2=530Pa{P}_{{SO}_{3}} = 43kPa; {P}_{{O}_{2}} = 530Pa and
PSO2=45kPa.{P}_{{SO}_{2}} = 45kPa . The equilibrium constant
{K}_{P} = _____ \times {10}^{- 2} . (Nearest integer)

Correct answer

0

Step-by-step solution

2SO2(g)+O2(g)=2SO3(g)2{SO}_{2 ( g )} + {O}_{2 ( g )} = 2{SO}_{3 ( g )}
KP=(pSO3(g))2(pSO2(g))2×pO2(g){K}_{P} = \dfrac{{\left({pSO}_{3 ( g )}\right)}^{2}}{{\left({{pSO}_{2}}_{\left(g\right)}\right)}^{2} \times {pO}_{2 ( g )}}
=43×4345×45×0.530= \dfrac{43 \times 43}{45 \times 45 \times 0 . 530}
=172.28×10−2kPa−1= 172 . 28 \times {10}^{- 2}{kPa}^{- 1}
≃172×10−2≃ 172 \times {10}^{- 2}
Ans is 172172
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