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Chemical Equilibrium question

2003 · Shift 0 · Q9
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Chemical Equilibrium question

2003 · Shift 0 · Q9

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
For the reaction equilibrium N2O4N_2O_4N2​O4​ (g) ⇋\leftrightharpoons⇋ 2NO2NO_2NO2​ (g) the concentrations of N2O4N_2O_4N2​O4​ and NO2NO_2NO2​ at equilibrium are 4.8 ×\times× 10-2 and 1.2 ×\times× 10-2 mol L-1 respectively. The value of Kc for the reaction is
  1. A
    3 ×\times× 10-1 mol L-1
  2. B
    3 ×\times× 10-3 mol L-1
  3. C
    3 ×\times× 103 mol L-1
  4. D
    3.3 ×\times× 102 mol L-1
View written solutionFree

Correct answer: B

  1. Write the equilibrium expression for N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g)N2​O4​(g)⇌2NO2​(g)

    The concentration equilibrium constant is: Kc=[NO2]2[N2O4]K_c = \frac{[NO_2]^2}{[N_2O_4]}Kc​=[N2​O4​][NO2​]2​

  2. Substitute the given equilibrium concentrations: [N2O4]=4.8×10−2 mol L−1[N_2O_4] = 4.8 \times 10^{-2}\,\text{mol L}^{-1}[N2​O4​]=4.8×10−2mol L−1 [NO2]=1.2×10−2 mol L−1[NO_2] = 1.2 \times 10^{-2}\,\text{mol L}^{-1}[NO2​]=1.2×10−2mol L−1

    So, Kc=(1.2×10−2)24.8×10−2K_c = \frac{(1.2 \times 10^{-2})^2}{4.8 \times 10^{-2}}Kc​=4.8×10−2(1.2×10−2)2​

  3. Calculate the numerator: (1.2×10−2)2=1.44×10−4(1.2 \times 10^{-2})^2 = 1.44 \times 10^{-4}(1.2×10−2)2=1.44×10−4

    Therefore, Kc=1.44×10−44.8×10−2K_c = \frac{1.44 \times 10^{-4}}{4.8 \times 10^{-2}}Kc​=4.8×10−21.44×10−4​

  4. Simplify: Kc=(1.444.8)×10−4+2K_c = \left(\frac{1.44}{4.8}\right) \times 10^{-4+2}Kc​=(4.81.44​)×10−4+2 Kc=0.3×10−2K_c = 0.3 \times 10^{-2}Kc​=0.3×10−2 Kc=3×10−3K_c = 3 \times 10^{-3}Kc​=3×10−3

  5. Unit check: Since Kc=(mol L−1)2mol L−1=mol L−1K_c = \frac{(\text{mol L}^{-1})^2}{\text{mol L}^{-1}} = \text{mol L}^{-1}Kc​=mol L−1(mol L−1)2​=mol L−1

    Hence, Kc=3×10−3 mol L−1K_c = 3 \times 10^{-3}\,\text{mol L}^{-1}Kc​=3×10−3mol L−1

  6. Match with the options: This corresponds to Option B.

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