Difficult
The following equilibrium constants are given: $N_2 + 3 H_2 \rightleftharpoons 2 NH_3$ $(k_1)$,$N_2 + O_2 \rightleftharpoons 2 NO$ $(k_2)$,$H_2 + 1/2 O_2 \rightleftharpoons H_2 O$ $(k_3)$. The equilibrium constant for the oxidation of $1 \text{ mole } NH_3$ by oxygen to give $NO$ according to the reaction $NH_3 + 5/4 O_2 \rightleftharpoons NO + 3/2 H_2 O$ is:
- A$\frac{k_2^{1/2} k_3^{3/2}}{k_1^{1/2}}$
- B$\frac{k_2^2 k_3}{k_1}$
- C$\frac{k_1 k_2}{k_3}$
- D$\frac{k_2 k_3^3}{k_1}$
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Similar Questions
If an equilibrium process equation is multiplied by a factor $n$,how does the equilibrium constant change? Explain with an example.
Medium
View SolutionIf the volume of the container is $1 \ L$ and at equilibrium the amounts are $SO_3 = 48 \ g$,$SO_2 = 12.8 \ g$,and $O_2 = 9.6 \ g$,find the value of $K_c$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.
Difficult
View SolutionDihydrogen gas is obtained from water gas by the following equation:
$\underbrace{CO_{(g)} + H_2O_{(g)}}_{water\,gas} \rightleftharpoons_{500^{\circ}C} CO_{2_{(g)}} + H_{2_{(g)}}$
At $733 \ K$,the concentrations of $CO$,$H_2O$,$CO_2$,and $H_2$ in water gas are $0.18$,$0.0412$,$0.15$,and $0.2 \ mol \ L^{-1}$ respectively. Find $K_c$.
$\underbrace{CO_{(g)} + H_2O_{(g)}}_{water\,gas} \rightleftharpoons_{500^{\circ}C} CO_{2_{(g)}} + H_{2_{(g)}}$
At $733 \ K$,the concentrations of $CO$,$H_2O$,$CO_2$,and $H_2$ in water gas are $0.18$,$0.0412$,$0.15$,and $0.2 \ mol \ L^{-1}$ respectively. Find $K_c$.
Medium
View SolutionGiven the reaction between $2$ gases represented by $A_2$ and $B_2$ to give the compound $AB_{(g)}$:
$A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$
At equilibrium,the concentrations are $[A_2] = 3.0 \times 10^{-3} \, M$,$[B_2] = 4.2 \times 10^{-3} \, M$,and $[AB] = 2.8 \times 10^{-3} \, M$.
If the reaction takes place in a sealed vessel at $527^{\circ}C$,then the value of $K_c$ will be:
$A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$
At equilibrium,the concentrations are $[A_2] = 3.0 \times 10^{-3} \, M$,$[B_2] = 4.2 \times 10^{-3} \, M$,and $[AB] = 2.8 \times 10^{-3} \, M$.
If the reaction takes place in a sealed vessel at $527^{\circ}C$,then the value of $K_c$ will be:
MediumAIPMT 2012
View Solution$3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$
For the above reaction at $298 \ K$,$K_c$ is found to be $3.0 \times 10^{-59}$. If the concentration of $O_2$ at equilibrium is $0.040 \ M$,then the concentration of $O_3$ in $M$ is ...... .
For the above reaction at $298 \ K$,$K_c$ is found to be $3.0 \times 10^{-59}$. If the concentration of $O_2$ at equilibrium is $0.040 \ M$,then the concentration of $O_3$ in $M$ is ...... .
MediumNEET 2022
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