JEE Main 2024 (Online) 4th April Evening Shift | Chemical Equilibrium Question 6 | Chemistry

JEE Main 2024 (Online) 4th April Evening Shift

MCQ (Single Correct Answer)

-1

The equilibrium constant for the reaction

$$\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$$

is $$\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}$$. The value of $$\mathrm{K}_{\mathrm{c}}$$ for the reaction given below is $$2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$$ is :

416

41.6

4.9

JEE Main 2024 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

-1

$$\mathrm{A}_{(\mathrm{g})} \rightleftharpoons \mathrm{B}_{(\mathrm{g})}+\frac{\mathrm{C}}{2}(\mathrm{g})$$ The correct relationship between $$\mathrm{K}_{\mathrm{P}}, \alpha$$ and equilibrium pressure $$\mathrm{P}$$ is

$$K_P=\frac{\alpha^{1 / 2} P^{3 / 2}}{(2+\alpha)^{3 / 2}}$$

$$K_P=\frac{\alpha^{3 / 2} P^{1 / 2}}{(2+\alpha)^{1 / 2}(1-\alpha)}$$

$$K_P=\frac{\alpha^{1 / 2} P^{1 / 2}}{(2+\alpha)^{3 / 2}}$$

$$K_P=\frac{\alpha^{1 / 2} P^{1 / 2}}{(2+\alpha)^{1 / 2}}$$

JEE Main 2024 (Online) 31st January Morning Shift

MCQ (Single Correct Answer)

-1

For the given reaction, choose the correct expression of $$\mathrm{K}_{\mathrm{C}}$$ from the following :-

$$\mathrm{Fe}_{(\mathrm{aq})}^{3+}+\mathrm{SCN}_{(\mathrm{aq})}^{-} \rightleftharpoons(\mathrm{FeSCN})_{(\mathrm{aq})}^{2+}$$

$$\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}{\left[\mathrm{FeSCN}^{2+}\right]}$$

$$\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}$$

$$\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]^2}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}$$

$$\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]^2\left[\mathrm{SCN}^{-}\right]^2}$$

JEE Main 2023 (Online) 6th April Morning Shift

MCQ (Single Correct Answer)

-1

For a concentrated solution of a weak electrolyte ($$\mathrm{K}_{\text {eq }}=$$ equilibrium constant) $$\mathrm{A}_{2} \mathrm{B}_{3}$$ of concentration '$$c$$', the degree of dissociation '$$\alpha$$' is :

$$\left(\frac{K_{e q}}{25 c^{2}}\right)^{\frac{1}{5}}$$

$$\left(\frac{K_{e q}}{108 c^{4}}\right)^{\frac{1}{5}}$$

$$\left(\frac{K_{e q}}{5 c^{4}}\right)^{\frac{1}{5}}$$

$$\left(\frac{K_{e q}}{6 c^{5}}\right)^{\frac{1}{5}}$$

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