Home
Class 12
CHEMISTRY
When NaNO(3) is heated in a closed vesse...

When `NaNO_(3)` is heated in a closed vessel, `O_(2)` is liberated and `NaNO_(2)` is left behind. At equilibrium: `NaNO_(3(g)) hArr NaNO_(2(g)) + (1)/(2) O_(2(g))`

A

addition of `NaNO_(2)` favours reverse reaction

B

addition of `NaNO_(3)` favours forward reaction

C

increasing temperature favours forward reaction

D

increasing pressure reduces the rate of forward reaction

Text Solution

Verified by Experts

The correct Answer is:
C, D
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    AAKASH SERIES|Exercise LECTURE SHEET( EXERCISE-II ) (LINKED COMPREHENSION TYPE QUESTIONS )|5 Videos

  • CHEMICAL EQUILIBRIUM

    AAKASH SERIES|Exercise LECTURE SHEET( EXERCISE-II ) (MATRIX MATCHING TYPE QUESTIONS )|2 Videos

  • CHEMICAL EQUILIBRIUM

    AAKASH SERIES|Exercise LECTURE SHEET( EXERCISE-II) (STRAIGHT OBJECTIVE TYPE QUESTIONS )|25 Videos

  • CARBOXYLIC ACIDS AND DERIVATIVES

    AAKASH SERIES|Exercise CONVERSIONS|19 Videos

  • CHEMICAL KINETCS

    AAKASH SERIES|Exercise EXERCISE - 3.2|45 Videos

Similar Questions

Explore conceptually related problems

1 mol A_((g)) is heated in 1 lit closed vessel and equilibrium is reached at 300^(@) C in A_((g)) hArr B_((g)) . If K_(C) =4 , concentration of B_((g)) at equilibrium is (in mol/lit)

At 627^(@) C and one atmosphere SO_(3) is is partially dissociated into SO_(2) and O_(2) by SO_(3(g)) hArr SO_(2(g)) + (1)/(2) O_(2(g)). The density of the equilibrium mixture is 0.925 g/litre. What is the degree of dissociation ?

18.4 g N_(2)O_(4) was placed in 1 L vessel at 400 K and allowed to attain the following equilibrium N_(2)O_(4)(g) hArr 2NO_(2)(g) . IF the total pressure at equilibrium was 10.64 bar, approximate K_(p) is (R = 0.083 L bar K^(-1) mol^(-1) ) (Assume N_(2)O_(4), NO_(2) as ideal gases)

What is the volume of oxygen liberated at S.T.P When 3.4g of H_(2)O_(2) is heated?

The equilibrium constant of the reaction, SO_(2)(g) + 1/2 O_(2)(g) hArr 2SO_(3)(g) is 5 xx 10^(-2)atm . The equilibrium constant of the reaction, 2SO_(3)(g) hArr 2SO_(2)(g) + O_(2)(g)

For the gaseous reactions (I) and (II) the equilibrium constants are X and Y respectively. I. (1)/(2)N_(2)(g) + O_(2) (g) hArr NO_(2)(g) (II) 2NO_(2)(g) hArr N_(2)O_(4) (g) Using the above reactions the equilibrium constant Z for the reaction (III) given below is III. N_(2)O_(4)(g) hArr N_(2)(g) + 2O_(2) (g)

The equilibrium constant for the given reaction is 100. N_(2)(g) + 2O_(2)(g) hArr 2NO_(2)(g) What is the equilibrium constant for the reaction given below? NO_(2)(g) hArr 1/2 N_(2)(g) + O_(2)

At constant temperature & volume , 50% of ozone is decomposed out of 2 atm of ozone taken initially and the equilibrium 2O_(3 (g)) hArr 3O_(2 (g)) is established , K_(p) for the decomposition of ozone is