AP EAMCET · Chemistry · Chemical Equilibrium
15 moles of \(\mathrm{H}_2\) and 5.2 moles of \(\mathrm{I}_2\) are mixed and allowed to attain equilibrium at 773 K . At equilibrium, the number of moles of HI is found to be 10 . The equilibrium constant for the dissociation of HI is
- A \(2 \times 10^{-2}\)
- B 50
- C \(2 \times 10^{-1}\)
- D 5
Answer & Solution
Correct Answer
(A) \(2 \times 10^{-2}\)
Step-by-step Solution
Detailed explanation
\begin{array}{llll} & \mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons & 2 HI \\Initial & 15 & 5.2 & 0 \\At equilibrium & 15-\mathrm{x} & 5.2-\mathrm{x} & 2 x\end{array} Let ' V ' is volume in liter Given, \(2 \mathrm{x}=10\) \(x=5\) Concentration of \(\mathrm{H}_2\) at equilibrium…
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