Home
Class 12
CHEMISTRY
On a given condition, the equilibrium co...

On a given condition, the equilibrium concentration of HI, `H_(2) "and" I_(2)` are 0.80, 0.10 and 0.10 `mol//L`. the equilibrium constant for the reaction, `H_(2) + I_(2) rarr 2HI` will be

A

64

B

12

C

8

D

0.8

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE - D|10 Videos

  • CHEMICAL EQUILIBRIUM

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE - E|10 Videos

  • CHEMICAL EQUILIBRIUM

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE - B|10 Videos

  • CHEMICAL BONDING-I & II

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|98 Videos

  • CHEMICAL KINETICS

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|52 Videos

Similar Questions

Explore conceptually related problems

On a given condition, the equilibrium concentration of HI , H_(2) and I_(2) are 0.80 , 0.10 and 0.10 mole/litre. The equilibrium constant for the reaction H_(2) + I_(2) hArr 2HI will be

Equilibrium concentration of HI, I_(2) and H_(2) is 0.7, 0.1 and 0.1 M respectively. The equilibrium constant for the reaction, I_(2)+H_(2)hArr 2HI is :

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

If the equilibrium constant of the reaction 2HIhArr H_(2)+I_(2) is 0.25, then the equilibrium constant for the reaction, H_(2)(g)+I_(2)(g)hArr 2HI(g) would be

Equilibrium concentration of HI,I2 and H2 is 0.7,0.1 and 0.1M respectively , Find Kc

The difference rate law for the reaction H_(2) +I_(2) to 2HI is

If the equilibrium constant of the reaction 2HI(g)hArrH_(2)(g)+I_(2)(g) is 0.25 , find the equilibrium constant of the reaction. (1)/(2)H_(2)+(1)/(2)I_(2)hArrHI(g)

The equilibrium constant for the reaction H_(2)(g)+I_(2)(g)hArr 2HI(g) is 32 at a given temperature. The equilibrium concentration of I_(2) and HI are 0.5xx10^(-3) and 8xx10^(-3)M respectively. The equilibrium concentration of H_(2) is

The value of equilibrium constant of the reaction. HI(g)hArr(1)/(2)H_2(g)+(1)/(2)I_2(g) is 8.0 The equilibrium constant of the reaction. H_2(g)+I_2(g)hArr2HI(g) will be

The value of the equilibrium constant for the reaction : H_(2) (g) +I_(2) (g) hArr 2HI (g) at 720 K is 48. What is the value of the equilibrium constant for the reaction : 1//2 H_(2)(g) + 1//2I_(2)(g) hArr HI (g)