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Chemical Equilibrium question

2021 · 17 Mar · Shift 2 · Q17
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Chemical Equilibrium question

2021 · 17 Mar · Shift 2 · Q17

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
Consider the reaction N2O4(g)⇌2NO2(g)N_{2}O_{4}\left( g\right) \rightleftharpoons 2NO_{2}\left( g\right)N2​O4​(g)⇌2NO2​(g) The temperature at which KC = 20.4 and KP = 600.1, is ‾\underline{\hspace{2cm}}​ K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K −-− 1 mol −-− 1]
Numerical answer
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Correct answer: 354

  1. For the reaction N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g)N2​O4​(g)⇌2NO2​(g) the relation between KPK_PKP​ and KCK_CKC​ is KP=KC(RT)ΔnK_P = K_C (RT)^{\Delta n}KP​=KC​(RT)Δn where Δn=(moles of gaseous products)−(moles of gaseous reactants)=2−1=1\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - 1 = 1Δn=(moles of gaseous products)−(moles of gaseous reactants)=2−1=1

  2. Hence, KP=KC(RT)K_P = K_C (RT)KP​=KC​(RT) Substituting the given values: 600.1=20.4×(0.0831)×T600.1 = 20.4 \times (0.0831) \times T600.1=20.4×(0.0831)×T

  3. Solve for TTT: T=600.120.4×0.0831T = \frac{600.1}{20.4 \times 0.0831}T=20.4×0.0831600.1​

  4. First calculate the denominator: 20.4×0.0831=1.6952420.4 \times 0.0831 = 1.6952420.4×0.0831=1.69524

  5. Therefore, T=600.11.69524≈354.0T = \frac{600.1}{1.69524} \approx 354.0T=1.69524600.1​≈354.0

  6. Rounded to the nearest integer, T=354 KT = 354\text{ K}T=354 K

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