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JEE Main 202429 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2024 · 29 January · Evening Shift · Chemistry · Chemical Equilibrium

The following concentrations were observed at 500K for the...

The following concentrations were observed at 500K500K for the formation of NH3{NH}_{3} from N2 andH2{N}_{2} and {H}_{2}. At equilibrium : \left[{N}_{2}\right] = 2 \times {10}^{- 2}M, \left[{H}_{2}\right] = 3 \times {10}^{- 2}Mand[NH3]=1.5×10−2Mand \left[{NH}_{3}\right] = 1 . 5 \times {10}^{- 2}M. Equilibrium constant for the reaction is ______.

Correct answer

0

Step-by-step solution

The equilibrium constant of concentration (denoted by Kc{K}_{c}) of a chemical reaction at equilibrium can be defined as the ratio of the concentration of products to the concentration of the reactants, each raised to their respective stoichiometric coefficients.

The equilibrium reaction is 

N2(g)+3H2(g)⇋2NH3(g){N}_{2\left(g\right)} + 3{H}_{2\left(g\right)}\mathop{⇋}_{}^{}2{NH}_{3\left(g\right)}

Kc=[NH3]2[H2]3[N2]{K}_{c} = \dfrac{{\left[{NH}_{3}\right]}^{2}}{{\left[{H}_{2}\right]}^{3}\left[{N}_{2}\right]}

=(32×10−2)2(3×10−2)3×2×10−2= \dfrac{{\left(\dfrac{3}{2} \times {10}^{- 2}\right)}^{2}}{{\left(3 \times {10}^{- 2}\right)}^{3} \times 2 \times {10}^{- 2}}

=32×10−422×33×10−6×2×10−2= \dfrac{{3}^{2} \times {10}^{- 4}}{{2}^{2} \times {3}^{3} \times {10}^{- 6} \times 2 \times {10}^{- 2}}

=14×2×3×104=1000024=416.6≈417= \dfrac{1}{4 \times 2 \times 3} \times {10}^{4} = \dfrac{10000}{24} = 416 . 6 \approx 417

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