164017 In an experiment the equilibrium constant for the reaction \(A+B \rightleftharpoons C+D\) is \(K\) when the initial concentration of \(A\) and \(B\) each is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\) Under the similar conditions in an another experiment if the initial concentration of \(A\) and \(B\) are taken 2 and \(3 \mathrm{~mol} \mathrm{~L}^{-1}\) respectively then the value of equilibrium constant will be:

164018 For the reaction: \(\mathbf{P} \rightleftharpoons \mathbf{Q}+\mathbf{R}\). Initially \(\mathbf{2} \mathbf{~ m o l}\) of \(\mathbf{P}\) was taken. Upto equilibrium \(0.5 \mathrm{~mol}\) of \(P\) was dissociated. What would be the degree of dissociation?

164019 \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)
In above reaction, at equilibrium condition mole fraction of \(\mathrm{PCl}_5\) is 0.4 and mole fraction of \(\mathrm{Cl}_2\) is 0.3. Then find out mole fraction of \(\mathrm{PCl}_3\)

164020 In \(A_3(g) \rightleftharpoons 3 A(g)\) reaction, the initial concentration of \(A_3\) is "a" \(\mathrm{mol} \mathrm{L}^{-1}\) If \(x\) is degree of dissociation of \(A_3\). The total number of moles at equilibrium will be:

164017 In an experiment the equilibrium constant for the reaction \(A+B \rightleftharpoons C+D\) is \(K\) when the initial concentration of \(A\) and \(B\) each is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\) Under the similar conditions in an another experiment if the initial concentration of \(A\) and \(B\) are taken 2 and \(3 \mathrm{~mol} \mathrm{~L}^{-1}\) respectively then the value of equilibrium constant will be:

164018 For the reaction: \(\mathbf{P} \rightleftharpoons \mathbf{Q}+\mathbf{R}\). Initially \(\mathbf{2} \mathbf{~ m o l}\) of \(\mathbf{P}\) was taken. Upto equilibrium \(0.5 \mathrm{~mol}\) of \(P\) was dissociated. What would be the degree of dissociation?

164019 \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)
In above reaction, at equilibrium condition mole fraction of \(\mathrm{PCl}_5\) is 0.4 and mole fraction of \(\mathrm{Cl}_2\) is 0.3. Then find out mole fraction of \(\mathrm{PCl}_3\)

164020 In \(A_3(g) \rightleftharpoons 3 A(g)\) reaction, the initial concentration of \(A_3\) is "a" \(\mathrm{mol} \mathrm{L}^{-1}\) If \(x\) is degree of dissociation of \(A_3\). The total number of moles at equilibrium will be:

164017 In an experiment the equilibrium constant for the reaction \(A+B \rightleftharpoons C+D\) is \(K\) when the initial concentration of \(A\) and \(B\) each is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\) Under the similar conditions in an another experiment if the initial concentration of \(A\) and \(B\) are taken 2 and \(3 \mathrm{~mol} \mathrm{~L}^{-1}\) respectively then the value of equilibrium constant will be:

164018 For the reaction: \(\mathbf{P} \rightleftharpoons \mathbf{Q}+\mathbf{R}\). Initially \(\mathbf{2} \mathbf{~ m o l}\) of \(\mathbf{P}\) was taken. Upto equilibrium \(0.5 \mathrm{~mol}\) of \(P\) was dissociated. What would be the degree of dissociation?

164019 \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)
In above reaction, at equilibrium condition mole fraction of \(\mathrm{PCl}_5\) is 0.4 and mole fraction of \(\mathrm{Cl}_2\) is 0.3. Then find out mole fraction of \(\mathrm{PCl}_3\)

164020 In \(A_3(g) \rightleftharpoons 3 A(g)\) reaction, the initial concentration of \(A_3\) is "a" \(\mathrm{mol} \mathrm{L}^{-1}\) If \(x\) is degree of dissociation of \(A_3\). The total number of moles at equilibrium will be:

164017 In an experiment the equilibrium constant for the reaction \(A+B \rightleftharpoons C+D\) is \(K\) when the initial concentration of \(A\) and \(B\) each is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\) Under the similar conditions in an another experiment if the initial concentration of \(A\) and \(B\) are taken 2 and \(3 \mathrm{~mol} \mathrm{~L}^{-1}\) respectively then the value of equilibrium constant will be:

164018 For the reaction: \(\mathbf{P} \rightleftharpoons \mathbf{Q}+\mathbf{R}\). Initially \(\mathbf{2} \mathbf{~ m o l}\) of \(\mathbf{P}\) was taken. Upto equilibrium \(0.5 \mathrm{~mol}\) of \(P\) was dissociated. What would be the degree of dissociation?

164019 \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)
In above reaction, at equilibrium condition mole fraction of \(\mathrm{PCl}_5\) is 0.4 and mole fraction of \(\mathrm{Cl}_2\) is 0.3. Then find out mole fraction of \(\mathrm{PCl}_3\)

164020 In \(A_3(g) \rightleftharpoons 3 A(g)\) reaction, the initial concentration of \(A_3\) is "a" \(\mathrm{mol} \mathrm{L}^{-1}\) If \(x\) is degree of dissociation of \(A_3\). The total number of moles at equilibrium will be: