AP EAMCET · Chemistry · Chemical Equilibrium
\(\mathrm{K}_{\mathrm{c}}\) for the reaction
\(\mathrm{A}_2(\mathrm{~g}) \stackrel{\mathrm{T}(\mathrm{~K})}{\rightleftharpoons} \mathrm{B}_2(\mathrm{~g})\)
is 39.0. In a closed one litre flask, one mole of \(\mathrm{A}_2(\mathrm{~g})\) was heated to \(\mathrm{T}(\mathrm{K})\). What are the concentrations of \(\mathrm{A}_2(\mathrm{~g})\) and \(\mathrm{B}_2(\mathrm{~g})\) (in \(\mathrm{mol}^{-1}\) ) respectively at equilibrium?
- A \(0.025,0.975\)
- B \(0.975,0.025\)
- C \(0.05,0.95\)
- D \(0.02,0.98\)
Answer & Solution
Correct Answer
(A) \(0.025,0.975\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Initial } \\ & \begin{array}{l}\text { Equilibrium } \\ \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{B}_2\right]}{\left[\mathrm{A}_2\right]} \\ 39=\frac{\left[\frac{\mathrm{x}}{1}\right]}{\left[\frac{1-\mathrm{x}}{1}\right]} \\ \text { or, } 39-39…
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