AP EAMCET · Chemistry · Chemical Equilibrium
\(\mathrm{K}_{\mathrm{c}}\) for the reaction, \(\mathrm{A}_2(\mathrm{~g}) \stackrel{\mathrm{T}(\mathrm{K})}{\rightleftarrows} \mathrm{B}_2(\mathrm{~g})\) is 99.0 . In a 1 L closed flask two moles of \(\mathrm{B}_2(\mathrm{~g})\) is heated to \(\mathrm{T}(\mathrm{K})\). What is the concentration of \(\mathrm{B}_2(\mathrm{~g})\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium?
- A \(0.02\)
- B \(1.98\)
- C \(0.198\)
- D \(1.5\)
Answer & Solution
Correct Answer
(B) \(1.98\)
Step-by-step Solution
Detailed explanation
Initial concentration of \(\mathrm{A}_2=0\) Initial concentration of \(\mathrm{B}_2=2\) Let x be the concentration of \(\mathrm{A}_2\) so the change in \(\mathrm{B}_2\) would be \(-2 x\) Equilibrium concentration are…
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