JEE Main 2024 (Online) 31st January Evening Shift | Chemical Equilibrium Question 7 | Chemistry

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 :

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

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

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

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

JEE Main 2022 (Online) 30th June Morning Shift

MCQ (Single Correct Answer)

-1

The equilibrium constant for the reversible reaction

2A(g) $$\rightleftharpoons$$ 2B(g) + C(g) is K₁

$${3 \over 2}$$A(g) $$\rightleftharpoons$$ $${3 \over 2}$$B(g) + $${3 \over 4}$$C(g) is K₂.

K₁ and K₂ are related as :

$${K_1} = \sqrt {{K_2}} $$

$${K_2} = \sqrt {{K_1}} $$

$${K_2} = K_1^{3/4}$$

$${K_1} = K_2^{3/4}$$

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