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JEE Main 201910 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2019 · 10 January · Morning Shift · Chemistry · Chemical Equilibrium

What are the values of K _ p K _ c for the following reactions at...

What are the values of KpKc\dfrac{{K}_{p}}{{K}_{c}} for the following reactions at 300K300K respectively?

(At 300K,RT=24.62dm2atmmol−1300K, RT = 24 .62 {dm}^{2}atm{mol}^{- 1} )

N2(g)+O2(g)⇌2NO(g){N}_{2}\left(g\right) + {O}_{2}\left(g\right) ⇌ 2NO\left(g\right)

N2O4(g)⇌2NO(g){N}_{2}{O}_{4}\left(g\right) ⇌ 2NO\left(g\right)

{N}_{2}\left(g\right) + 3{H}_{2}\left(g\right) ⇌ 2{NH}_{3}\left(g\right)

Options

  1. A24.63dm3atmmol−1,606.0dm6atm2mol−2,1.65×10−3dm−6atm−2mol2\begin{array}{ll}24 .63 {dm}^{3}atm{mol}^{- 1}, \\ 606 .0 {dm}^{6}{atm}^{2}{mol}^{- 2}, \\ 1 .65 \times {10}^{- 3}{dm}^{- 6}{atm}^{- 2}{mol}^{2}\end{array}
  2. B1,24.62dm6atm3mol−1606.0dm6atm2mol−2\begin{array}{ll}1 ,24.62 {dm}^{6}{atm}^{3}{mol}^{- 1} \\ 606 .0 {dm}^{6}{atm}^{2}{mol}^{- 2}\end{array}
  3. C1,24.62dm3atmmol−11.65×10−3dm−6atm−2mol2\begin{array}{ll}1 ,24.62 {dm}^{3}atm{mol}^{- 1} \\ 1 .65 \times {10}^{- 3}{dm}^{- 6}{atm}^{- 2}{mol}^{2}\end{array}
  4. D1,4.1×10−2dm−3atm−1mol,606dm6atm2mol−2\begin{array}{ll}1 , 4.1 \times {10}^{- 2}{dm}^{- 3}{atm}^{- 1}mol, \\ 606{dm}^{6}{atm}^{2}{mol}^{- 2}\end{array}

Correct answer

C. 1,24.62dm3atmmol−11.65×10−3dm−6atm−2mol2\begin{array}{ll}1 ,24.62 {dm}^{3}atm{mol}^{- 1} \\ 1 .65 \times {10}^{- 3}{dm}^{- 6}{atm}^{- 2}{mol}^{2}\end{array}

Step-by-step solution

∵KP=KC(RT)Δng∵ {K}_{P} = {K}_{C}{\left(RT\right)}^{\Delta ng}

KPKC=(RT)Δng\dfrac{{K}_{P}}{{K}_{C}} = {\left(RT\right)}^{\Delta ng}

(i) N2+O2⇌2NOΔng=0∴KPKC=1{N}_{2} + {O}_{2} ⇌ 2NO\Delta ng = 0 ∴ \dfrac{{K}_{P}}{{K}_{C}} = 1

(ii) N2O4⇌2Na2Δng=1∴KPKC=(RT)1=(24.62)1{N}_{2}{O}_{4} ⇌ 2{Na}_{2}\Delta ng = 1 ∴ \dfrac{{K}_{P}}{{K}_{C}} = {\left(RT\right)}^{1} = {\left(24 .62\right)}^{1}

(iii) N2+3H2⇌2NH3Δng=−2{N}_{2} + 3{H}_{2} ⇌ 2{NH}_{3}\Delta ng = - 2 ∴KPKC=(RT)−2=(24.62)−2=1.659×10−3∴ \dfrac{{K}_{P}}{{K}_{C}} = {\left(RT\right)}^{- 2} = {\left(24 .62\right)}^{- 2} = 1 .659 \times {10}^{- 3}≃1.65×10−3≃ 1.65 \times {10}^{- 3}
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