Question

Please answer 7 , 8 , 10 and 11 questions!


Lazy answers will be deleted on spot!

Expecting good answer!

Answer 1

this is the answer of question number 7 and 8

Answer 2

Oxidation numbers (usually same as valency)

7. a) H4 P2 O7  
    Oxidation numbers of  H= +1,  O = -2, Compound = 0 (neutral).
    Oxidation number of Phosphorus = x.
    4 * 1 + 2 * x - 7 * 2 = 0   =>   x = 5
   
b) Na B H4  =  Na⁺ & BH₄⁻
    Oxidation numbers:  Na = 1,   So oxidation number of BH₄ = -1
    Here Hydrogen has an oxidation number -1, as it gets one electron from Boron.
    Oxidation number of Boron = x.
    So  x - 4 * 1 = -1     =>   x = +3

c)  CrO5
     Oxidation number of O = -2.  Overall = 0.
     So  x + 5 * (-2) = 0    =>  x = 10
     Chromium has an oxid.num = +10.
===================
11.  Iron Oxide  Feₐ Oₓ   = Neutral compound
      "a" number of atoms of Iron combined with "x" atoms of Oxygen.
      Mass of the molecule = a * 56 + x * 16
      % of Fe in the compound by mass = 56a /(56a+16x) * 100
      Given  7a/(7a+2x) = 69.9/100    =>  30.1 * 7a = 2 * 69.9 x   ---(1)
      So a/x = 0.66 = 2/3  approximately

      So 3 a = 2 x    =>   a = 2, x = 3  are smallest integers satisfying it.
       So formula:    Fe₂ O₃
      It is the Ferric Oxide  or Iron (III) Oxide.
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Q8.  Equilibrium constants:
   2 NO Cl (g)  ⇔ 2 NO (g) + Cl₂ (g)
      2x moles          0                0               : at start
      2x (1- α)          2x α           xα moles    :  At Equilibrium
     [NO] = 2xα/V    ;   [Cl₂] = xα/V     ;    [NOCl ] = 2x(1-α)/V
  
   Let there be 2x moles  of NOCl at the start in the container. Let α be the dissociation constant of the reaction at the given temperature. Let V be the volume of the container.
    Kc = [ NO ]² [ Cl₂ ] / [ NOCl ]²
          = [4 x³ α³/V³] / [4x² (1-α)²/V² ]
          =  x α³ / [V (1-α)²] = 3.75 * 10⁻⁶ mole/l        ---- (1)

 Total number of moles n = 2x(1-α)+ 2xα + xα = x(2+α)
 Ideal gas law:    P V = n R T
 Total pressure in the container = P = x (2+α) R T/V,   T = 1069°K
  Partial pressures:   p_NO = molefraction * total pressure
           p_NO = 2xα P / [x(2+α)] = 2xα RT/V
           p_Cl₂ = x α P / [x(2+α)] =  xα RT/V
           p_NOCL = 2x(1-α) P / [x(2+α)] = 2x(1-α) RT/V

=>  Kp = [p_NO]² [ p_Cl₂ ] / [ p_NOCl ]²
           = [ 4x³α³ R³ T³ / V³ ] / [ 4 x² (1-α)² R²T² /V² ]
           = x α³ R T / [ V (1 - α)² ]         ------ (2)
           = Kc * R T        using (1)
           = 3.75 * 10⁻⁶ mol/l * 8.314 J/mol/°K  * 1069°K
           = 3.333 * 10⁻² J/liter

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Q10.  Equilibrium constant (pressure)  Kp = ?   Hydrogen Iodide.

At the equilibrium, it is given
    Total pressure P = 0.2 atm
    Partial pressure P_HI = 0.04 atm
    Molar ratio of HI = P_HI / P = 0.04/0.2 = 1/5

       2 H I (g)     ⇔   H₂ (g) +   I₂ (g)

   Molar ratio of HI is 1/5 at equilibrium. Also the number of moles of H2 and I2 are equal at the equilibrium (from the equation).

=> So molar ratio of H2 = [ 1 - 1/5 ] ÷ 2 = 2/5
=>                         of I2 = 2/5
=> Partial pressure of H2 = P_H2 = 2/5 * P
=> Partial pressure of I2 = P_I2 = 2/5 * P

Equilibrium constant = Kp = [ p_H₂ ] [ p_I₂ ] / [ p_HI ]²
        Kp = [ 2/5 P * 2/5 P ] / [ 1/5 P ]²
             = 4

Answer is   Kp = 4    (no units)