JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
At \(-20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure, a cylinder is filled with equal number of \(\mathrm{H}_2 . \mathrm{I}_2\) and \(\mathrm{HI}\) molecules for the reaction \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\), the \(\mathrm{K}_{\mathrm{p}}\) for the process is \(\mathrm{x} \times 10^{-1} \cdot \mathrm{x}=\) _______. [Given : \(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) ]
- A \(2\)
- B \(1\)
- C \(10\)
- D \(0.01\)
Answer & Solution
Correct Answer
(C) \(10\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{n}_{\mathrm{g}}=0 \mathrm{~K}_{\mathrm{p}}=\frac{\left(\mathrm{n}_{\mathrm{HI}}\right)^2}{\mathrm{n}_{\mathrm{H}_2} \mathrm{n}_{\mathrm{I}_2}}\left(\frac{\mathrm{P}_{\mathrm{T}}}{\mathrm{n}_{\mathrm{T}}}\right)^{\Delta \mathrm{ng}_{\mathrm{g}}}\)…
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