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NEET-UG and JEE Main Chemistry · Class 11

Law of chemical equilibrium; Kp and Kc — NEET & JEE Questions

Law of chemical equilibrium; Kp and Kc is a topic in Equilibrium, a Class 11 Chemistry chapter of the NEET-UG and JEE Main syllabus. TestBudy holds 84 published NEET-UG and JEE Main questions on Law of chemical equilibrium; Kp and Kc, each with the marking scheme of the real paper.

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Law of chemical equilibrium; Kp and Kc MCQs with solutions

10 of the 84 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L1 · recall

    For the reaction aA+bB⇌cC+dDaA+bB\rightleftharpoons cC+dD, the concentration equilibrium constant is

    • A. Kc=[C]c[D]d[A]a[B]bK_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}
    • B. Kc=[A]a[B]b[C]c[D]dK_c=\frac{[A]^a[B]^b}{[C]^c[D]^d}
    • C. Kc=[A]+[B]+[C]+[D]K_c=[A]+[B]+[C]+[D]
    • D. Kc=abcdK_c=abcd
  2. Question 2 · difficulty L2 · understanding

    One mole each of He and A(g)\mathrm{A}(\mathrm{g}) are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium. A( g)⇌B( g) \mathrm{A}(\mathrm{~g}) \rightleftharpoons \mathrm{B}(\mathrm{~g}) Kc\mathrm{K}_{\mathrm{c}} for this reaction at 400 K is 4.0 . The partial pressures (in atm) of He and B(g) are respectively (at equilibrium) (Assume He,A(g)\mathrm{He}, \mathrm{A}(\mathrm{g}) and B(g)\mathrm{B}(\mathrm{g}) behave as ideal gases) (Given : R=0.082 L atm K−1 mol−1\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} )

    • A. 3.28, 2.624
    • B. 2.624, 3.28
    • C. 3.28, 0.656
    • D. 0.656, 6.56
  3. Question 3 · difficulty L2 · understanding

    At T(K)\mathrm{T}(\mathrm{K}), the equilibrium constant of A2(g)+B2(g)⇌C(g)\mathrm{A}_2(g)+\mathrm{B}_2(g) \rightleftharpoons \mathrm{C}(g) is 2.7×10−52.7 \times 10^{-5}. What is the equilibrium constant for 13 A2( g)+13 B2( g)⇌13C(g)\frac{1}{3} \mathrm{~A}_2(\mathrm{~g})+\frac{1}{3} \mathrm{~B}_2(\mathrm{~g}) \rightleftharpoons \frac{1}{3} \mathrm{C}(\mathrm{g}) at the same temperature?

    • A. (2.7×10−5)3\left(2.7 \times 10^{-5}\right)^3
    • B. 6×10−26 \times 10^{-2}
    • C. 2.7×10−5\sqrt{2.7 \times 10^{-5}}
    • D. 3×10−2 3\times 10^{-2}
  4. Question 4 · difficulty L2 · understanding

    Given below are two statements : Statement I : A catalyst cannot alter the equilibrium constant (Kc)\left(\mathrm{K}_{\mathrm{c}}\right) of the reaction, temperature remaining constant. Statement II : A homogenous catalyst can change the equilibrium composition of a system, temperature remaining constant. In the light of the above statements, choose the correct answer from the options given below

    • A. Statement I is true but Statement II is false
    • B. Statement I is false but Statement II is true
    • C. Both Statement I and Statement II are true
    • D. Both Statement I and Statement II are false
  5. Question 5 · difficulty L2 · understanding

    A vessel at 1000 K contains CO2\mathrm{CO}_2 with a pressure of 0.5 atm . Some of CO2\mathrm{CO}_2 is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm , then K p is :

    • A. 0.18 atm
    • B. 0.3 atm
    • C. 3 atm
    • D. 1.8 atm
  6. Question 6 · difficulty L2 · understanding

    For the given hypothetical reactions, the equilibrium constants are as follows : X⇌Y;K1=1.0Y⇌Z;K2=2.0Z⇌W;K3=4.0\begin{aligned} & \mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0 \\ & \mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0 \\ & \mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0 \end{aligned} The equilibrium constant for the reaction X⇌W\mathrm{X} \rightleftharpoons \mathrm{W} is

    • A. 12.0
    • B. 8.0
    • C. 6.0
    • D. 7.0
  7. Question 7 · difficulty L2 · understanding

    The ratio KPKC\frac{K_P}{K_C} for the reaction : CO(g)+12O2( g)⇌CO2( g)\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})} is :

    • A. 1
    • B. (RT)1/2 (\mathrm{RT})^{1 / 2}
    • C. RT
    • D. 1RT\mathrm{ \frac{1}{\sqrt{R T}}}
  8. Question 8 · difficulty L2 · understanding

    At −20∘C-20^{\circ} \mathrm{C} and 1 atm1 \mathrm{~atm} pressure, a cylinder is filled with equal number of H2,I2\mathrm{H}_2, \mathrm{I}_2 and HI\mathrm{HI} molecules for the reaction H2( g)+I2( g)⇌2HI(g)\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}), the Kp\mathrm{K}_{\mathrm{p}} for the process is x×10−1x \times 10^{-1}. x=\mathrm{x}= __________. [Given : R=0.082 L atm K−1 mol−1\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}]

    • A. 2
    • B. 1
    • C. 10
    • D. 0.01
  9. Question 9 · difficulty L2 · understanding

    The equilibrium constant for the reaction SO3( g)⇌SO2( g)+12O2( g)\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) is Kc=4.9×10−2\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}. The value of Kc\mathrm{K}_{\mathrm{c}} for the reaction given below is 2SO2( g)+O2( g)⇌2SO3( g)2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) is :

    • A. 49
    • B. 416
    • C. 41.6
    • D. 4.9
  10. Question 10 · difficulty L2 · understanding

    For the given reaction, choose the correct expression of KC\mathrm{K}_{\mathrm{C}} from the following :- Fe(aq)3++SCN(aq)−⇌(FeSCN)(aq)2+\mathrm{Fe}_{(\mathrm{aq})}^{3+}+\mathrm{SCN}_{(\mathrm{aq})}^{-} \rightleftharpoons(\mathrm{FeSCN})_{(\mathrm{aq})}^{2+}

    • A. KC=[Fe3+][SCN−][FeSCN2+]\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}{\left[\mathrm{FeSCN}^{2+}\right]}
    • B. KC=[FeSCN2+][Fe3+][SCN−]\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}
    • C. KC=[FeSCN2+]2[Fe3+][SCN−]\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]^2}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}
    • D. KC=[FeSCN2+][Fe3+]2[SCN−]2\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]^2\left[\mathrm{SCN}^{-}\right]^2}

Common mistakes in Law of chemical equilibrium; Kp and Kc

Recorded against questions in this topic that are not shown above, so none of these points at an answer you can see on this page.

  • Uses coefficients as multiplicative factors rather than exponents.
  • Forgets the exponent $\Delta n_g$.
  • Assumes an RT factor remains even when the gaseous mole change is zero.

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