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Equilibrium Test - 49

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Equilibrium Test - 49
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  • Question 1
    1 / -0
    The thermal dissociation equilibrium of CaCO3(s)CaC{O}_{3}\left(s\right) is studied under different conditions.
                 CaCO3(s)CaO(s)+CO2(g)CaC{O}_{3}\left(s\right) \rightleftharpoons CaO\left(s\right) + C{O}_{2}\left(g\right)
    For this equilibrium, the correct statement(s) is (are) :
    Solution
    Option (C)(C) is correct.
    KK is dependent on the pressure of CO2C{O}_{2} and CaCO3CaCO_3 at a given TT.
    When a system at equilibrium undergoes a change in pressure, the equilibrium of the system will shift to offset the change and establish a new equilibrium. The system can shift in one of two ways:

    Toward the reactants (i.e. in favour of the reverse reaction)
    Toward the products (i.e. in favour of the forward reaction)
    The effects of changes in pressure can be described as follows (this only applies to reactions involving gases):
    When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with fewer moles of gas.
    When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of gas.
  • Question 2
    1 / -0
    Assertion: On mixing equal volumes of 1 M HClHCl and of 2 M CH3COONaCH_3COONa, an acidic buffer solution is formed.
    Reason: The resultant mixture contains CH3COOHCH_3COOH and CH3COONaCH_3COONa which are parts of acidic buffer.
    Solution
    A solution that resists change in pH value upon addition of small amount of strong acid or base (less than 1 %) or when solution is diluted is called buffer solution.
    Acidic buffer solution : An acidic buffer solution consists of solution of a weak acid and its salt with strong base. The best-known example is a mixture of solution of acetic acid and its salt with strong base (CH3COONa)(CH_3COONa).
    Reaction:
    CH3COONa+HClCH3COONa+CH3COOHCH_3COONa + HCl \rightarrow CH_3COONa + CH_3COOH 
         2               1                        -                     -
        1                   -                                         1                          1
    which are parts of acidic buffer
  • Question 3
    1 / -0
    The reaction which proceeds in the forward direction is:
    Solution

  • Question 4
    1 / -0

    Directions For Questions

    The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion of ammonia molecules reacts with H2O{H}_{2}O to form NH4OH{NH}_{4}OH molecules. 

    NH4OH{NH}_{4}OH again dissociates into NH4+{NH}_{4}^{+} and OH{OH}^{-} ions. In solution therefore, we have NH3{NH}_{3} molecules, NH4OH{NH}_{4}OH molecules and NH4+{NH}_{4}^{+} ions and the following equilibrium exist:

    NH3(g){NH}_{3}(g) NH3(l)+H2O(l) NH4OH(aq) NH4+(aq)+OH(aq) \rightleftharpoons{NH}_{3}(l)+{H}_{2}O{(l)}\rightleftharpoons  {NH}_{4}OH{(aq)}\rightleftharpoons  {NH}_{4}^{+}{(aq)}+{OH}^{-}{(aq)}

    Let, c1{c}_{1} (mol/L){(mol/L)} of NH3{NH}_{3} pass in liquid state which on dissolution in water forms c2{c}_{2} (mol/L){(mol/L)} of NH4OH{NH}_{4}OH. The solution contains c3{c}_{3} (mol/L){(mol/L)} of NH4+{NH}_{4}^{+} ions

    ...view full instructions

    The dissociation constant of NH4OH{NH}_{4}OH can be given as
    Solution
    As given,

    NH4OHc2(c2c3)NH4+0c3+OH0c3\underset { { c }_{ 2 }\\ ({ c }_{ 2 }-{ c }_{ 3 }) }{ { NH }_{ 4 }OH } \rightleftharpoons \underset { 0\\ { c }_{ 3 } }{ { NH }_{ 4 }^{ + } } +\underset { 0\\ { c }_{ 3 } }{ { OH }^{ - } }

    Kb=[NH4+][OH]/[NH4OH]K_b = [NH_4^+][OH^-]/[NH_4OH]

    Kb=c32(c2c3)  \quad { K }_{ b }=\cfrac { { { c }_{ 3 } }^{ 2 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right)  }
  • Question 5
    1 / -0
    The equilibrium constant for the reaction, N2(g)+O2(g)2NO(g){N}_{2}\left(g\right) + {O}_{2}\left(g\right) \rightleftharpoons 2NO\left(g\right) is 4×1044\times {10}^{-4} at 2000 K2000  K. In presence of a catalyst, the equilibrium is attained 1010 times faster. Therefore, the equilibrium constant in the presence of the catalyst at 2000K2000 \:K is:
    Solution
    The equilibrium constant in the presence of the catalyst at 2000K2000\:K is 4×1044\times{10}^{-4}.

    Kp{K}_{p} and Kc{K}_{c} values do not change with catalyst. They only change with temperature. 
    With catalyst equilibrium is attained faster.

    Hence, the value of equilibrium constant remains the same.
  • Question 6
    1 / -0
    Given a reaction:
    Cl2(g)+3F2(g)2ClF3(g);  ΔH=329 kJC{l}_{2}\left(g\right) + 3{F}_{2}\left(g\right) \rightleftharpoons 2Cl{F}_{3}\left(g\right);    \Delta H = -329  kJ
    Which of the following will increase the quantity of ClF3Cl{F}_{3} in an equilibrium mixture of Cl2,F2C{l}_{2}, {F}_{2} and ClF3Cl{F}_{3}?
    Solution
    Favourable conditions for forward reaction according to Le Chatelier's principle are the decrease in temperature, increase in concentration of reactants and increase in pressure.
    Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to re-establish an equilibrium.
    If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change.
    Le Chatelier's principle implies that the addition of heat to a reaction will favor the endothermic direction of a reaction as this reduces the amount of heat produced in the system.
    Increasing the concentration of reactants will drive the reaction to the right, while increasing the concentration of products will drive the reaction to the left.
    Hence, addition of F2F_2  will increase the quantity of ClF3Cl{F}_{3} in an equilibrium mixture of Cl2,F2C{l}_{2}, {F}_{2} and ClF3Cl{F}_{3}.
  • Question 7
    1 / -0
    The volume of the reaction vessel containing an equilibrium mixture in the reaction,

                   SO2Cl2(g)SO2(g)+Cl2(g)S{O}_{2}C{l}_{2}\left(g\right) \rightleftharpoons S{O}_{2}\left(g\right) + C{l}_{2}\left(g\right)

    is increased. When equilibrium is re-established : 
    Solution
    An increase in volume for the equilibrium having,

                    $${K}_{c} = \displaystyle\frac{\left[S{O}_{2}\right]\left[C{l}_{2}\right]}{\left[S{O}_{2}C{l}_{2}\right]} = \displaystyle\frac{\left[mole\  of\  S{O}_{2}\right]\left[mole\  of\  C{l}_{2}\right]}{\left[mole\  of\  S{O}_{2}C{l}_{2}\right]\times Volume}$$

    will increase the mole of Cl2C{l}_{2} or SO2S{O}_{2} to maintain Kc{K}_{c} values constant.
  • Question 8
    1 / -0
    Densities of diamond and graphite are 3.53.5 and 2.3 g/mL2.3  {g}/{mL} respectively. Increase of pressure on the equilibrium:

    C(diamond)C(graphite){C}_{\left(diamond\right)} \rightleftharpoons {C}_{\left(graphite\right)}
    Solution
    According to Le Chatlier's principle, if the pressure increases at equilibrium the reaction occurs in the direction where the volume decreases which is nothing but the density increased side.

    As the density of the diamond is more than the graphite, the backward direction is favourable.

    Hence option A is correct. 
  • Question 9
    1 / -0
    Acetyl salicylic acid (aspirin) ionises in water as: HC9H7O4+H2OH3O++C9H7O4;HC_9H_7O_4+H_2O\rightarrow H_3O^+ + C_9H_7O_4^-; (Ka=2.75×109)(K_a = 2.75\times 10^{-9}). If two tablets of aspirin each of 0.32 g is dissolved in water to produce 250 mL solution, calculate [OH].[\overset{\circleddash}{O}H]. 
    Solution
    [HC9H7O4]=0.32×2×1000180×250=0.014M[HC_9H_7O_4] = \displaystyle \frac{0.32\times 2\times 1000}{180\times 250} = 0.014\, M

     α=KaC=2.75×1090.014=4.43×104\because   \alpha = \displaystyle \sqrt{\frac{K_a}{C}} = \sqrt{\frac{2.75\times 10^{-9}}{0.014}} = 4.43\times 10^{-4}

     [H]=C.α=0.014×4.43×104=6.21×106M\therefore  [H^{\oplus}] = C.\alpha = 0.014 \times 4.43\times 10^{-4}=6.21\times 10^{-6}\, M

         [C9H7O4]=C.α=6.21×106M[C_9H_7O_4^{\ominus}]=C.\alpha = 6.21\times 10^{-6}\, M

         [OH]=1014[H]=10146.21×106=1.61×109M[OH^{\circleddash}]=\displaystyle \frac{10^{-14}}{[H^{\oplus}]}=\frac{10^{-14}}{6.21\times 10^{-6}}= 1.61\times 10^{-9}M
  • Question 10
    1 / -0
    Calculate the equilibrium constants for the reactions with water of H2PO4,HPO42H_2PO_4^{\circleddash}, HPO_4^{2-} and PO43PO_4^{3-} as base. Comparing the relative values of two equilibrium constants of H2PO4H_2PO_4^{\circleddash} with water, deduce whether solutions of this ion in water are acidic or bases. Deduce whether solutions of HPO42HPO_4^{2-} are acidic or bases. Given K1,K2K_1, K_2 and K3K_3 for H3PO4H_3PO_4 are 7.1×103,6.3×1087.1\times 10^{-3}, 6.3\times 10^{-8} and 4.5×10134.5\times 10^{-13} respectively .
    Solution
    i. H2PO4+H2O  H3PO4+OHH_2PO_4^{\circleddash} + H_2O \rightleftharpoons  H_3PO_4+\overset{\circleddash}{O}H
       Kh=KwKa=10147.1×103=1.4×1012K_h = \displaystyle \frac{K_w}{K_a}=\frac{10^{-14}}{7.1\times 10^{-3}}=1.4\times 10^{-12}
    ii. HPO42+H2O H2PO4+OHHPO_4^{2-} + H_2O \rightleftharpoons H_2PO_4^{\circleddash}+\overset{\circleddash}{O}H
       Kh=KwK2=10146.3×108=1.6×107K_h = \displaystyle \frac{K_w}{K_2} = \frac{10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}
    iii. PO43+H2O HPO42+OHPO_4^{3-}+H_2O \rightleftharpoons HPO_4^{2-}+\overset{\circleddash}{O}H
       Kh=KwK3=10144.5×1013=2.2×102K_h = \displaystyle \frac{K_w}{K_3} = \frac{10^{-14}}{4.5\times 10^{-13}}=2.2\times 10^{-2}
    H2PO4H_2PO_4^{\circleddash} can react with water to yield HPO42HPO_4^{2-} and H3OH_3O^{\oplus} with a constant 6.3×1086.3\times 10^{-8}
    Its reaction with water to yield H3PO4H_3PO_4, OH\overset{\circleddash}{O}H has a much smaller equilibrium constant, 1.4×10121.4 \times 10^{-12}, and so this ion in water is acidic comparision of the constants for HPO42(4.5×103HPO_4^{2-}(4.5\times 10^{-3} as an acid, 1.6×1071.6\times 10^{-7} as base )) leads to the conclusion that this ion in solution is basic.
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