The artificial fixation of nitrogen (N2) has enormous energy, environmental, and societal impact, the most important of which is the synthesis of ammonia (NH3) for fertilizers that help support nearly half of the world's population.
<h3>Artificial fixation of nitrogen</h3>
a) The equilibrium constant expression is Kp=PCH4 PH2 OP CO×PH 23.
(b) (i) As the pressure increases, the equilibrium will shift to the left so that less number of moles are produced.
(ii) For an exothermic reaction, with the increase in temperature, the equilibrium will shift in the backward direction.
(iii) When a catalyst is used, the equilibrium is not disturbed. The equilibrium is quickly attained
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In the physical properties, it is mentioned that the element has 4 valence electrons. The elements in the periodic table are arranged such that, the number of valence electrons present in the neutral atoms belonging to a particular group is equal to the group number.
Thus, the unidentified element can be best classified as a nonmetal in period 4
Ans C)
When a solid (solute) comes in contact with the liquid (solvent), the solute goes about C) dissolution, in which the solid dissolves into the liquid.
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Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
