AP EAPCET 2024 - 18th May Morning Shift | Chemical Equilibrium Question 5 | Chemistry

At $T(\mathrm{~K}), K_c$ for the reaction $A_2(g) \rightleftharpoons B_2(g)$ is 99.0 . Two moles of $A_2(s)$ was heated to $T(\mathrm{~K})$ in a 1 L . closed flask to reach the above equilibrium. What are the concentrations (in mol $\mathrm{L}^{-1}$ ) of $A_2(g)$ and $B_2(g)$ respectively at equilibrium?

$1,86,0.0187$

$1.98,0.02$

$0.0187,1,86$

$0.02,1.98$

AP EAPCET 2024 - 18th May Morning Shift

MCQ (Single Correct Answer)

-0

At $27^{\circ} \mathrm{C}$, the degree of dissociation of weak acid ( $\mathrm{H} A$ ) in its 0.5 M aqueous solution is $1 \%$. Its $K_e$, value is approximately

$5 \times 10^{-6}$

$5 \times 10^{-5}$

$5 \times 10^{-6}$

$5 \times 10^{-8}$

AP EAPCET 2022 - 5th July Morning Shift

MCQ (Single Correct Answer)

-0

At 500 K , for the reaction $$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$$, the $$K_p$$ is $$0.036 \mathrm{~atm}^{-2}$$. What is its $$K_C$$ in $$\mathrm{L}^2 \mathrm{~mol}^{-1}$$ ? $$\left(R=0.082 \mathrm{~L}^2\right.$$ atom $$\left.\mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)$$.

$$2.1 \times 10^{-4}$$

$$2.1 \times 10^{-5}$$

60.5

605

AP EAPCET 2022 - 4th July Morning Shift

MCQ (Single Correct Answer)

-0

The formation of ammonia from its constituent elements is an exothermic reaction. The effect of increased temperature on the reaction equilibrium is

the rate of the forward reaction becomes zero.

no effect of temperature.

forward reaction is favoured

backward reaction is favoured

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