Chemical Equilibrium · Chemistry · COMEDK

For a reaction,

$$A+B \rightleftharpoons 2 C$$

1.0 mole of $A, 1.5$ mole of $B$ and 0.5 mole of $C$ were taken in a 1 L vessel. At equilibrium, the concentration of $C$ was $1.0 \mathrm{~mol} \mathrm{~L}^{-1}$. The equilibrium constant for the reaction is $x / 15$. The value of ' $x$ ' is:

Consider the reaction

$$\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+3 \mathrm{CO}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Fe}(\mathrm{l})+3 \mathrm{CQ}_2(\mathrm{~g})$$

In accordance with Le-Chatlier's principle, which of the following will not disturbs the equilibrium?

At $$700 \mathrm{~K}$$, the Equilibrium constant value for the formation of $$\mathrm{HI}$$ from $$\mathrm{H}_2$$ and $$\mathrm{I}_2$$ is 49.0 . 0.7 mole of $$\mathrm{HI}(\mathrm{g})$$ is present at equilibrium. What will be the concentrations of $$\mathrm{H}_2$$ and $$\mathrm{I}_2$$ gases if we initially started with $$\mathrm{HI}(\mathrm{g})$$ and allowed the reaction to reach equilibrium at the same temperature?

$$\mathrm{S}_8$$ on heating at a temperature above $$1000 \mathrm{~K}$$, changes to $$\mathrm{S}_2$$. When 1 mole of $$\mathrm{S}_8$$ is heated above $$1000 \mathrm{~K}$$, the pressure falls by $$32 \%$$ at equilibrium. The equilibrium constant for the conversion is:

Consider the following equilibrium,

$$\begin{aligned} & 2 \mathrm{No}(g) \rightleftharpoons \mathrm{N}_2+\mathrm{O}_2 ; \mathrm{K}_{\mathrm{G}}=2.4 \times 10^{20} \\ & \mathrm{No}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NoBr}(\mathrm{g}) ; \mathrm{K}_{\mathrm{C}_2}=1.4 \end{aligned}$$

Calculate $$K_C$$ for the reaction,

$$\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g)+\frac{1}{2} \mathrm{Br}_2(g) \rightleftharpoons \mathrm{NOBr}(g)$$

The equilibrium constants for the reactions $$a, b$$, and $$c$$ are as given:

a) $$\mathrm{N}_2+3 \mathrm{H}_2=2 \mathrm{NH}_3: \mathbf{K}_1$$

b) $$\mathrm{N}_2+\mathrm{O}_2=2 \mathrm{NO}: \mathrm{K}_2$$

c) $$2 \mathrm{H}_2+\mathrm{O}_2=2 \mathrm{H}_2 \mathrm{O}: \mathbf{K}_3$$

What would be the Equilibrium constant for the reaction:

$$4 \mathrm{NH}_3+5 \mathrm{O}_2=4 \mathrm{NO}+6 \mathrm{H}_2 \mathrm{O} ; \mathbf{K}_{\mathbf{x}}$$

For the reaction, H$$_2$$(g) + I$$_2$$(g) $$\rightleftharpoons$$ 2HI(g) the position of equilibrium can be shifted to the right by

The equilibrium constant, $$K_C$$ for $$3C_2H_2(g) \rightleftharpoons C_6H_6(g)$$ is 4 L$$^2$$mol$$^{-2}$$. If the equilibrium concentration of benzene is 0.5 mol$$^{-1}$$ than what is the value of concentration of ethylene?

In the equilibrium, $$AB\rightleftharpoons A+B$$, if the equilibrium concentration of A is double, then equilibrium concentration of B will be

For the equilibrium,

2NOCl(g) $$\rightleftharpoons$$ 2NO(g) + Cl$$_2$$(g),

the value of the equilibrium constant, $$K_C$$ is 3.75 $$\times$$ 10$$^{-6}$$ at 1069 K. The value of $$K_p$$ for the reaction at this temperature will be

In the chemical reaction,

N$$_2$$ + 3H$$_2$$ $$\rightleftharpoons$$ 2NH$$_3$$ at equilibrium point.

$$2S{O_2}(g) + {O_2}(g)\buildrel {{V_2}{O_5}} \over \rightleftharpoons $$ is an example for