Consider the reaction `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` for which `K_(c)=278M^(-1)`.0.001 mole ofeach of the reagents `SO_(2)(g),O_(2)(g)and SO_(3)(g)` are mixed in a 1.0 L flask . Dterminr=e the reaction quotient of the system and the spontaneus direction of the system:
A
`Q_(c)=1000,`the equilibrium shifts to the right
B
`Q_(c)=1000,`the equilibrium shifts to the left
C
`Q_(c)=0.001,` the equilibrium shifts to the left
D
`Q_(c)=0.001, ` theequilibrium shifts to the right
Text Solution
Verified by Experts
The correct Answer is:
a
Topper's Solved these Questions
CHEMICAL EQUILIBRIUM
NARENDRA AWASTHI|Exercise Level 1 (Q.93 To Q.122)|1 Videos
CHEMICAL EQUILIBRIUM
NARENDRA AWASTHI|Exercise Level 2|1 Videos
ATOMIC STUCTURE
NARENDRA AWASTHI|Exercise level 2|1 Videos
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
Similar Questions
Explore conceptually related problems
For a reaction N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))
For the reactions : C(s) + O_(2) (g) to CO_(2) (g)
Write the reactions of SO_(2) with : K_(2)Cr_(2)O_(7) .
Consider the reaction 2CO(g)+O_(2)(g)hArr2CO_(2)(g)+Heat Under what conditions shift is undetminable ?
For the reaction CO(g)+Cl_(2)(g)hArrCOCl_(2)(g) the value of (K_(c)/(K_(P))) is equal to :
For the reaction H_(2)(g)+I_(2)(g)hArr2HI(g) the equilibrium constant K_(p) changes with
The equilibrium constant for the reaction N_(2)(g)+O_(2)(g) hArr 2NO(g) at temperature T is 4xx10^(-4) .The value of K_(c) for the reaction NO(g) hArr 1/2 N_(2)(g)+1/2 O_(2)(g) at the same temperature is
For the reaction 2NO_(2)(g)+(1)/(2)O_(2)(g)hArrN_(2)O_(5)(g) if the equilibrium constant is K_(p) , then the equilibrium constant for the reaction 2N_(2)O_(5)(g)hArr4NO_(2)(g)+O_(2)(g) would be :
The equilibrium constant K_(c) for the reaction SO_(2) (g) + NO_(2)(g)hArrSO_(3)(g)+NO(g) is 16 . 1 mole of each of all the four gases is taken in 1dm^(3) vessel , the equilibrium concentration of NO would be:
For the first order reaction 2N_2O_5 (g) rarr 4 NO_2(g)+ O_2(g) which is incorrect:-
