AP EAMCET · Chemistry · Chemical Equilibrium
One mole of \(\mathrm{A}(\mathrm{g})\) is heated to \(\mathrm{T}(\mathrm{K})\) till the following equilibrium is obtained
\(\mathrm{A}(\mathrm{g}) \stackrel{\mathrm{T}(\mathrm{K})}{\rightleftharpoons} \mathrm{B}(\mathrm{g})\)
The equilibrium constant of this reaction is \(10^{-1}\). After reaching the equilibrium, 0.5 moles of \(\mathrm{A}(\mathrm{g})\) is added and heated. The equilibrium is again established. The value of \(\frac{[\mathrm{A}]}{[\mathrm{B}]}\) is
- A \(10^{-1}\)
- B 10
- C \(10^{-2}\)
- D 100
Answer & Solution
Correct Answer
(B) 10
Step-by-step Solution
Detailed explanation
\(\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{B}]}{[\mathrm{A}]}=10^{-1}\) or \(\frac{1}{10}\) Since \(\mathrm{K}_{\mathrm{C}}\) remains constant at a given temperature, the value of \(\frac{[\mathrm{B}]}{[\mathrm{A}]}\) will still be the same.…
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