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N(2)O(4(g))rArr2NO(2),K(c)5.7xx10^(-9) ...

`N_(2)O_(4(g))rArr2NO_(2),K_(c)5.7xx10^(-9)` at 298 K At equilibrium :-

A

concentration of `NO_(2)` is higher than that of `N_(2)O_(4)`

B

concentration of `N_(2)O_(4)` is higher than of `NO_(2)`

C

both `N_(2)O_(4) and NO_(2)` have same concentration

D

concentration of `N_(2)O_(4) and NO_(2)` keeps on changing

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2
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At a certain temperature , the following reactions have the equilibrium constants as shown below: S(s)+O_(2)(g)hArrSO_(2)(g),K_(c)=5xx10^(52) 2S(s)+3O_(2)(g)hArr2SO_(3)(g),K_(c)=5xx10^(29) what is the equilibrium constant K_(c) for the reaction at tahea same temperature? 2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)

At 298 K , the equilibrium between N_(2)O_(4) and NO_(2) is represented as : N_(2)O_(4(g))hArr2NO_(2(g)) . If the total pressure of the equilibrium mixture is P and the degree of dissociation of N_(2)O_(4(g)) at 298 K is x , the partial pressure of NO_(2(g)) under these conditions is:

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In a 1 lit. Container following equilibrium is estabilished with equal moles of NO_(2)(g) & N_(2)O_(4)(g) . N_(2)O_(4)(g) hArr 2NO_(2)(g) hArr 2NO_(2)(g) at equilibrium M_(avg.)=(184)/(3) , then ratio of K_(c) & total initial mole is .

18.4 g of N_(2)O_(4) is taken in a 1 L closed vessel and heated till the equilibrium is reached. N_(2)O_(4(g))rArr2NO_(2(g)) At equilibrium it is found that 50% of N_(2)O_(4) is dissociated . What will be the value of equilibrium constant?

Consider the reaction, NO_(2) rarr 1/2N_(2) + O_(2),K_(1) , N_(2)O_(4) rarr 2NO_(2) , K_(2) Give the equilibrium constant for the formation of N_(2)O_(4) "from" N_(2) "and" O_(2) .

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