TS EAMCET · Chemistry · Chemical Equilibrium
Consider the equilibrium, \(\mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}\). Calculate the equilibrium constant of the reverse reaction when the equilibrium concentration of \(\mathrm{H}_2, \mathrm{I}_2\) and \(\mathrm{HI}\) are \(1.14 \times 10^{-2}, 0.12 \times 10^{-2}\) and \(2.52 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}\), respectively
- A 46.4
- B 0.021
- C 18.42
- D 0.054
Answer & Solution
Correct Answer
(B) 0.021
Step-by-step Solution
Detailed explanation
\(\mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}\) Reverse of above reaction is \(2 \mathrm{HI} \rightleftharpoons \mathrm{H}_2+\mathrm{I}_2\) Equilibrium constant \((K)=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}\) Given, Concentration…
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