Easy
The equilibrium constant for the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$ is $5 \times 10^{-2} \ atm^{-1/2}$. The equilibrium constant of the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$ would be
- A$400 \ atm$
- B$200 \ atm$
- C$4 \times 10^2 \ atm$
- D$6.25 \times 10^4 \ atm$
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Similar Questions
The value of $K_{C}$ is $64$ at $800 \ K$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
The value of $K_{C}$ for the following reaction is:
$NH_{3(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)}$
The value of $K_{C}$ for the following reaction is:
$NH_{3(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)}$
State the law of chemical equilibrium for an equilibrium process.
Difficult
View SolutionThe equilibrium constants for the following reactions are given at $25^{\circ} C$:
$2 A \rightleftharpoons B + C, K_{1} = 1.0$
$2 B \rightleftharpoons C + D, K_{2} = 16$
$2 C + D \rightleftharpoons 2 P, K_{3} = 25$
The equilibrium constant for the reaction $P \rightleftharpoons A + \frac{1}{2} B$ at $25^{\circ} C$ is
$2 A \rightleftharpoons B + C, K_{1} = 1.0$
$2 B \rightleftharpoons C + D, K_{2} = 16$
$2 C + D \rightleftharpoons 2 P, K_{3} = 25$
The equilibrium constant for the reaction $P \rightleftharpoons A + \frac{1}{2} B$ at $25^{\circ} C$ is
MediumWBJEE 2020
View SolutionAssertion: Reaction quotient is defined in the same way as equilibrium constant at any stage of the reaction.
Reason: If $Q_c < K_c$,the reaction moves in the direction of reactants.
Reason: If $Q_c < K_c$,the reaction moves in the direction of reactants.
MediumAIIMS 2011
View SolutionIn the reaction $AB_{(g)} + CD_{(g)} \rightleftharpoons AD_{(g)} + CB_{(g)}$,one mole of $AB$ reacts with one mole of $CD$. When equilibrium is established,$3/4$ mole of each $AB$ and $CD$ are converted into $AD$ and $CB$. If the volume remains constant,the equilibrium constant for the reaction will be........
Medium
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