AP EAMCET · Chemistry · Chemical Equilibrium
\(\mathrm{K}_{\mathrm{c}}\) for the following reaction is 99.0
\(\mathrm{A}_2(\mathrm{~g}) \stackrel{\mathrm{T}(\mathrm{~K})}{\rightleftharpoons} \mathrm{B}_2(\mathrm{~g})\)
In a one litre flask, 2 moles of \(\mathrm{A}_2\) was heated to \(\mathrm{T}(\mathrm{K})\) and the above equilibrium is reached. The concentrations at equilibrium of \(\mathrm{A}_2\) and \(\mathrm{B}_2\) are \(\mathrm{C}_1\left(\mathrm{~A}_2\right)\) and \(\mathrm{C}_2\left(\mathrm{~B}_2\right)\) respectively. Now, one mole of \(\mathrm{A}_2\) was added to flask and heated to \(\mathrm{T}(\mathrm{K})\) to establish the equilibrium again. The concentrations of \(\mathrm{A}_2\) and \(\mathrm{B}_2\) are \(\mathrm{C}_3\left(\mathrm{~A}_2\right)\) and \(\mathrm{C}_4\left(\mathrm{~B}_2\right)\) respectively. What is the value of \(\mathrm{C}_3\left(\mathrm{~A}_2\right)\) in \(\mathrm{mol}^{-1}\) ?
- A 1.98
- B 0.01
- C 0.03
- D 2.97
Answer & Solution
Correct Answer
(C) 0.03
Step-by-step Solution
Detailed explanation
Given, \(\mathrm{K}_{\mathrm{c}}=99.0\) On adding one mole of \(\mathrm{A}_2\) Hence, \(\mathrm{C}_3\left(\mathrm{~A}_2\right)=0.03\) moles.
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