Question:
At \(1\,\mathrm{K}\), \(15\) moles of \(\mathrm{H_2}\) reacts with \(5.2\) moles of \(\mathrm{I_2}\) and forms \(10\) moles of \(\mathrm{HI}\). The equilibrium constant for the reaction
\[
2\mathrm{HI}(g)\rightleftharpoons \mathrm{H_2}(g)+\mathrm{I_2}(g)
\]
is
At \(1\,\mathrm{K}\), \(15\) moles of \(\mathrm{H_2}\) reacts with \(5.2\) moles of \(\mathrm{I_2}\) and forms \(10\) moles of \(\mathrm{HI}\). The equilibrium constant for the reaction
\[
2\mathrm{HI}(g)\rightleftharpoons \mathrm{H_2}(g)+\mathrm{I_2}(g)
\]
is
Show Hint
Use an ICE (Initial--Change--Equilibrium) table for equilibrium problems. If all species are in the same volume, moles may be used directly in the equilibrium expression because the volume terms cancel.
Updated On: Jul 9, 2026
- \(50\)
- \(100\)
- \(2\times10^{-2}\)
- \(1\times10^{-2}\) \bigskip
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
For the reaction
\[
2\mathrm{HI}\rightleftharpoons \mathrm{H_2}+\mathrm{I_2},
\]
the equilibrium constant is
\[
K_c=\frac{[\mathrm{H_2}][\mathrm{I_2}]}{[\mathrm{HI}]^2}.
\]
Step 1: Determine equilibrium moles. Initially, \[ \mathrm{H_2}=15,\qquad \mathrm{I_2}=5.2,\qquad \mathrm{HI}=0 \] Since \(10\) moles of HI are formed, \[ \mathrm{H_2+I_2\rightarrow 2HI} \] Therefore, \[ x=\frac{10}{2}=5 \] Equilibrium moles: \[ \mathrm{H_2}=15-5=10 \] \[ \mathrm{I_2}=5.2-5=0.2 \] \[ \mathrm{HI}=10 \]
Step 2: Calculate \(K_c\). \[ K_c=\frac{10\times0.2}{10^2} =\frac{2}{100} =2\times10^{-2} \]
Step 3: Final conclusion. \[ \boxed{K_c=2\times10^{-2}} \] Hence, the correct option is \(\boxed{(C)}\).
Step 1: Determine equilibrium moles. Initially, \[ \mathrm{H_2}=15,\qquad \mathrm{I_2}=5.2,\qquad \mathrm{HI}=0 \] Since \(10\) moles of HI are formed, \[ \mathrm{H_2+I_2\rightarrow 2HI} \] Therefore, \[ x=\frac{10}{2}=5 \] Equilibrium moles: \[ \mathrm{H_2}=15-5=10 \] \[ \mathrm{I_2}=5.2-5=0.2 \] \[ \mathrm{HI}=10 \]
Step 2: Calculate \(K_c\). \[ K_c=\frac{10\times0.2}{10^2} =\frac{2}{100} =2\times10^{-2} \]
Step 3: Final conclusion. \[ \boxed{K_c=2\times10^{-2}} \] Hence, the correct option is \(\boxed{(C)}\).
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