Skip to main content
IIT JEE Test Series — Practice smarter, perform stronger.

Chemical Equilibrium flash cards

Master Chemical Equilibrium through 91 JEE Advanced-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Chemical Equilibrium, question and answer

19 of this chapter's 91 cards, laid out open so you can read straight through. The remaining 72 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.What is a reversible reaction? Give an example.

    A reaction that proceeds in both forward and backward directions under the same conditions, so reactants are never fully consumed. Example: N2+3H22NH3\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3.

    Hint: Both arrows, never goes to completion.

  2. 2.Define dynamic equilibrium.

    The state in a reversible reaction where the forward and backward reaction rates become equal, so concentrations of all species stay constant with time even though both reactions continue. It is 'dynamic' because reactions never stop.

    Hint: Rates equal, concentrations constant, but activity continues.

  3. 3.List the characteristics of chemical equilibrium.

    Attained only in a closed system; dynamic in nature (rates equal, not reactions stopped); reachable from either direction; concentrations of reactants and products stay constant; a catalyst does not shift it; ΔG=0\Delta G = 0 at equilibrium.

    Hint: Closed system, dynamic, both directions, ΔG=0\Delta G=0.

  4. 4.State the Law of Mass Action.

    At constant temperature, the rate of a reaction is proportional to the product of the active masses (molar concentrations) of the reactants, each raised to the power of its stoichiometric coefficient.

    Hint: Rate \propto product of active masses raised to coefficients.

  5. 5.For aA+bBcC+dDa\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}, write the equilibrium constant KcK_c.

    Kc=[C]c[D]d[A]a[B]bK_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}, where the brackets denote equilibrium molar concentrations.

    Hint: Products over reactants, powers = coefficients.

  6. 6.Distinguish KcK_c and KpK_p.

    KcK_c is the equilibrium constant expressed using molar concentrations, while KpK_p is expressed using partial pressures of gaseous species. KpK_p is defined only when gases are involved.

    Hint: Concentration vs partial pressure.

  7. 7.Derive/state the relation between KpK_p and KcK_c.

    Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}, where Δn=(moles of gaseous products)(moles of gaseous reactants)\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) and RR is in units consistent with pressure and concentration.

    Hint: Use PV=nRTPV=nRT, so P=(n/V)RT=CRTP=(n/V)RT=CRT.

  8. 8.For which reactions is Kp=KcK_p = K_c?

    When Δng=0\Delta n_g = 0, i.e. equal moles of gas on both sides, since (RT)0=1(RT)^0 = 1. Example: H2+I22HI\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}.

    Hint: Δng=0(RT)0\Delta n_g = 0 \Rightarrow (RT)^0.

  9. 9.Are KcK_c and KpK_p dimensionless? Explain.

    Thermodynamically, KK is defined using activities (or ratios to standard states), making it dimensionless. In JEE numericals, KcK_c and KpK_p carry apparent units of (mol L1)Δn(\text{mol L}^{-1})^{\Delta n} and (atm)Δn(\text{atm})^{\Delta n} respectively unless Δn=0\Delta n = 0.

    Hint: Rigorously dimensionless; apparent units from Δn\Delta n.

  10. 10.What is the equilibrium constant in terms of mole fraction, KxK_x, and how does it relate to KpK_p?

    Kx=xiνi(products)xiνi(reactants)K_x = \dfrac{\prod x_i^{\nu_i}(\text{products})}{\prod x_i^{\nu_i}(\text{reactants})} and Kp=KxPΔnK_p = K_x \cdot P^{\Delta n}, where PP is total pressure. Thus KxK_x depends on pressure unless Δn=0\Delta n = 0.

    Hint: Kp=KxPΔnK_p = K_x P^{\Delta n}.

  11. 11.How does the equilibrium constant change when a reaction is reversed?

    The equilibrium constant of the reversed reaction is the reciprocal: K=1/KK' = 1/K.

    Hint: Reverse \Rightarrow invert.

  12. 12.How does KK change if all coefficients of a reaction are multiplied by nn?

    The new equilibrium constant becomes KnK^n. For example, doubling coefficients gives K2K^2.

    Hint: Multiply coefficients by nn \Rightarrow raise KK to power nn.

  13. 13.If reaction 1 (K1K_1) and reaction 2 (K2K_2) are added to give reaction 3, what is K3K_3?

    For added reactions, equilibrium constants multiply: K3=K1×K2K_3 = K_1 \times K_2.

    Hint: Add reactions \Rightarrow multiply KK's.

  14. 14.Define the reaction quotient QQ and how it differs from KK.

    QQ has the same algebraic form as KK but uses concentrations (or pressures) at any arbitrary instant, not necessarily equilibrium. At equilibrium Q=KQ = K.

    Hint: Same form as KK, any-instant values.

  15. 15.How is the direction of a reaction predicted using QQ and KK?

    If Q<KQ < K, forward reaction dominates (more products form); if Q>KQ > K, backward reaction dominates; if Q=KQ = K, the system is at equilibrium.

    Hint: Q<KQ<K forward, Q>KQ>K backward.

  16. 16.For a reaction, Q=25Q = 25 and K=100K = 100. Which direction does it proceed?

    Since Q<KQ < K, the reaction proceeds forward (to the right) to form more products until QQ rises to 100100.

    Hint: Compare QQ with KK; Q<KQ<K.

  17. 17.What does the magnitude of KK tell you about the extent of reaction?

    K1K \gg 1 means the equilibrium lies far to the right (products favoured); K1K \ll 1 means it lies to the left (reactants favoured); K1K \approx 1 means comparable amounts of both.

    Hint: Large KK = product-favoured.

  18. 18.Define degree of dissociation α\alpha.

    The fraction of one mole of a substance that has dissociated at equilibrium: α=moles dissociatedinitial moles\alpha = \dfrac{\text{moles dissociated}}{\text{initial moles}}. It lies between 0 and 1 (or expressed as a percentage).

    Hint: Fraction dissociated per mole.

  19. 19.For PCl5PCl3+Cl2\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 starting with 1 mol at total pressure PP, express KpK_p in terms of α\alpha.

    Total moles =1+α=1+\alpha. Kp=α21α2PK_p = \dfrac{\alpha^2}{1-\alpha^2}\,P, since each product mole fraction is α1+α\frac{\alpha}{1+\alpha} and reactant 1α1+α\frac{1-\alpha}{1+\alpha}.

    Hint: Kp=α21α2PK_p=\frac{\alpha^2}{1-\alpha^2}P for 1\to2 dissociation.

Open the interactive deck for the other 72 cards, with self-grading so the ones you keep missing come back.

Other ways to revise this chapter

Master this chapter with similar other learning materials.

Preparing students for India’s top institutes

Our students are currently into top technological and medical institutes of India.

  • IIT Bombay
  • IIT Delhi
  • IIT Madras
  • IIT Kanpur
  • IIT Kharagpur
  • IIT Roorkee
  • IIT Guwahati
  • IIT BHU Varanasi
  • AIIMS Delhi
  • NIT Tiruchirappalli
  • NIT Rourkela

Join QuestPix, Today!

Get notified first, with exam & curriculum updates, course & test series launch offers, motivation & success stories and free learning resources recommended by toppers.

Chat on WhatsApp