NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Chemical Equilibrium
NTA Abhyas 2020 · Chemistry · Chemical Equilibrium
Consider the following gaseous equilibria with equilibrium...
Consider the following gaseous equilibria with equilibrium constants {\text{K}_{1}\text{a}\text{n}\text{d}{\text{K}_{2} respectively.
\text{S}{\text{O}_{2}\left(\text{g}\right) + \dfrac{1}{2}{\text{O}_{2} ( \text{g} ) ⇌ \text{S}{\text{O}_{3} ( \text{g} )
2{\text{S}\text{O}_{3}\left(\text{g}\right) ⇌ 2{\text{S}\text{O}_{2}\left(\text{g}\right) + {\text{O}_{2} ( \text{g} )
The equilibrium constants are related as
\text{S}{\text{O}_{2}\left(\text{g}\right) + \dfrac{1}{2}{\text{O}_{2} ( \text{g} ) ⇌ \text{S}{\text{O}_{3} ( \text{g} )
2{\text{S}\text{O}_{3}\left(\text{g}\right) ⇌ 2{\text{S}\text{O}_{2}\left(\text{g}\right) + {\text{O}_{2} ( \text{g} )
The equilibrium constants are related as
Options
- A2{\text{K}_{1} = {\text{K}_{2}^{2}
- B{\text{K}_{1}^{2} = \dfrac{1}{{\text{K}_{2}}
- C{\text{K}_{2}^{2} = \dfrac{1}{{\text{K}_{1}}
- D{\text{K}_{2} = \dfrac{2}{{\text{K}_{1}^{2}}
Correct answer
B. {\text{K}_{1}^{2} = \dfrac{1}{{\text{K}_{2}}
Step-by-step solution
For the reaction,
\text{S}{\text{O}_{2}\left(\text{g}\right) + \dfrac{1}{2}{\text{O}_{2} ( \text{g} ) ⇌ \text{S}{\text{O}_{3} ( \text{g} )
Equilibrium constant, {\text{K}_{1} = \dfrac{[ \text{S}{\text{O}_{3} ]}{\left[\text{S}{\text{O}_{2}\right]{\left[{\text{O}_{2}\right]}^{1 / 2}}...(i)
For the reaction,
2{\text{S}\text{O}_{3}\left(\text{g}\right) ⇌ 2{\text{S}\text{O}_{2}\left(\text{g}\right) + {\text{O}_{2} ( \text{g} )
Equilibrium constant, {\text{K}_{2} = \dfrac{{\left[\text{S}{\text{O}_{2}\right]}^{2} [ {\text{O}_{2} ]}{{\left[\text{S}{\text{O}_{3}\right]}^{2}}...(ii)
On squaring both sides in Eq.(i), we get
{\text{K}_{1}^{2} = \dfrac{{\left[\text{S}{\text{O}_{3}\right]}^{2}}{{\left[\text{S}{\text{O}_{2}\right]}^{2} [ {\text{O}_{2} ]}...(iii)
Eq. (ii) Eq (iii), we get
{\text{K}_{1}^{2} \times {\text{K}_{2} = 1
or {\text{K}_{2} = \dfrac{1}{{\text{K}_{1}^{2}}
or {\text{K}_{1}^{2} = \dfrac{1}{{\text{K}_{2}}
\text{S}{\text{O}_{2}\left(\text{g}\right) + \dfrac{1}{2}{\text{O}_{2} ( \text{g} ) ⇌ \text{S}{\text{O}_{3} ( \text{g} )
Equilibrium constant, {\text{K}_{1} = \dfrac{[ \text{S}{\text{O}_{3} ]}{\left[\text{S}{\text{O}_{2}\right]{\left[{\text{O}_{2}\right]}^{1 / 2}}...(i)
For the reaction,
2{\text{S}\text{O}_{3}\left(\text{g}\right) ⇌ 2{\text{S}\text{O}_{2}\left(\text{g}\right) + {\text{O}_{2} ( \text{g} )
Equilibrium constant, {\text{K}_{2} = \dfrac{{\left[\text{S}{\text{O}_{2}\right]}^{2} [ {\text{O}_{2} ]}{{\left[\text{S}{\text{O}_{3}\right]}^{2}}...(ii)
On squaring both sides in Eq.(i), we get
{\text{K}_{1}^{2} = \dfrac{{\left[\text{S}{\text{O}_{3}\right]}^{2}}{{\left[\text{S}{\text{O}_{2}\right]}^{2} [ {\text{O}_{2} ]}...(iii)
Eq. (ii) Eq (iii), we get
{\text{K}_{1}^{2} \times {\text{K}_{2} = 1
or {\text{K}_{2} = \dfrac{1}{{\text{K}_{1}^{2}}
or {\text{K}_{1}^{2} = \dfrac{1}{{\text{K}_{2}}