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Chemical Equilibrium Test - 4

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Chemical Equilibrium Test - 4
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Which of the following is not a general characteristic of equilibria involving physical processes?

    Solution

    The physical processes do not stop at equilibrium. There is a state of dynamic equilibrium in these processes.

     

  • Question 2
    1 / -0

    Which of the following conditions represent(s) an equilibrium?

    Solution

    Equilibrium can be achieved only in closed vessel (the balloon in this case).

     

  • Question 3
    1 / -0

    Which of the following statements is incorrect?

    Solution

    The statement is incorrect. The reaction taking place in solution is:
    Fe3+ (aq) + 3SCN- (aq) ⇌ Fe(SCN)3
    Oxalic acid (H2C2O4) reacts with Fe3+ (aq) present in solution to form the complex Fe(SCN)3. As a result, the concentration of Fe3+ ions in the solution decreases and the intensity of red colour decreases.

     

  • Question 4
    1 / -0

    A 50 ml solution of 0.1 M AgNO3 is mixed with 50 ml of 0.01 M NaBrO3 solution. Ksp of AgBrO3 is 5.8 x 10-5. Which of the following is correct?

    Solution

    Precipitation takes place when ionic product is more than solubility product.
    Initial concentration of Ag+ ions = 0.1 M
    Initial concentration of BrO3- ions = 0.01 M
    Final volume of solution = 50 + 50 = 100 ml
    Final [Ag+] = 0.1 × 50/100 = 0.05
    Final [BrO3-] = 0.01 × 50/100 = 0.005
    Product of ionic concentrations of AgBrO3 = [Ag+] [BrO3-] = 0.05 × 0.005 = 2.5 × 10-4, which is greater than the solubility product.
    Hence, precipitate of AgBrO3 will be formed.

     

  • Question 5
    1 / -0

    At 700 K, the equilibrium constant Kp for the reaction 2SO3 (g) ⇌ 2SO(g) + O2 is 1.8 x 10–3 kPa. What is the numerical value in moles per liter of Kc for the reaction at the same temperature?

    Solution

    Kp = Kc (RT)∆n
    Kc = Kp/(RT)∆n
    T = 700 K
    R = 8.314 J mol-1K-1
    ∆n = (2 + 1) – 2 = 1
    Kp = 1.8 X 10-3kPa = 1.8 Pa
    Therefore,
    Kc = 1.8 Pa / (8.314 J mol-1 K-1) (700)1
    = 3.09 X 10-7 mol m-3
    Hence, the numerical value in moles per liter of Kc is equal to 3.09 X 10-7 mol m-3.

     

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