Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
EasyWhich of the following statements is false?
- AThe greater the concentration of the substances involved in a reaction,the lower the speed of the reaction.
- BThe point of dynamic equilibrium is reached when the reaction rate in one direction just balances the reaction rate in the opposite direction.
- CThe dissociation of a weak electrolyte is a reversible reaction.
- DThe presence of free ions facilitates chemical changes.
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Similar Questions
Consider the following gas phase reaction being carried out in a closed vessel at $25^\circ\text{C}$: $2A(g) \rightarrow 4B(g) + C(g)$. The table provides the total pressure of the system at different time intervals. Calculate the pressure of $C(g)$ at $30$ minutes time interval.
Time (min) Total pressure (mm Hg) $30$ $300$ $\infty$ $600$
| Time (min) | Total pressure (mm Hg) |
| $30$ | $300$ |
| $\infty$ | $600$ |
DifficultJEE MAIN 2026
View Solution$5 \ \text{moles}$ of $SO_2$ and $5 \ \text{moles}$ of $O_2$ are allowed to react to form $SO_3$ in a closed vessel. At the equilibrium stage,$60\%$ of $SO_2$ is used up. The total number of moles of $SO_2$,$O_2$,and $SO_3$ in the vessel now is:
Difficult
View SolutionFor the equilibrium $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ in a closed vessel at a constant temperature,if the volume of the reaction vessel is halved,which of the following statements is true regarding the equilibrium constant $K_p$ and the degree of dissociation $(\alpha)$?
Medium
View SolutionFor the reaction,$A \rightleftharpoons n B$,the concentration of $A$ decreases from $0.06 \ mol \ L^{-1}$ to $0.03 \ mol \ L^{-1}$ and that of $B$ rises from $0$ to $0.06 \ mol \ L^{-1}$ at equilibrium. The values of $n$ and the equilibrium constant for the reaction,respectively,are
MediumKVPY 2013
View SolutionThe equilibrium constants of the following are
$N_2 + 3H_2 \rightleftharpoons 2NH_3 \,; \quad K_1$
$N_2 + O_2 \rightleftharpoons 2NO \,; \quad K_2$
$H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \,; \quad K_3$
The equilibrium constant $(K)$ of the reaction:
$2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O$ is:
$N_2 + 3H_2 \rightleftharpoons 2NH_3 \,; \quad K_1$
$N_2 + O_2 \rightleftharpoons 2NO \,; \quad K_2$
$H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \,; \quad K_3$
The equilibrium constant $(K)$ of the reaction:
$2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O$ is:
DifficultAIPMT 2003
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