Home
Class 12
CHEMISTRY
H(2) O (I) hArr H(2) O(g) " "II) I(2)...

`H_(2) O (I) hArr H_(2) O(g) " "II) I_(2)(s) hArr I_(2)` (vap) `" "` III) `H_(2) O(l) hArr H_(2) O `(s) `" IV)" CO_(2) (g) hArr CO_(2)` (aq) Rise of T shifts equilibrium towards right in the case of

A

I & IV

B

II, III & IV

C

I & II

D

I, II & III

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of determining which equilibria shift to the right with an increase in temperature, we need to analyze each given equilibrium reaction based on whether they are endothermic or exothermic. ### Step-by-Step Solution: 1. **Identify the Equilibria**: - **I)** \( H_2O (l) \rightleftharpoons H_2O (g) \) - **II)** \( I_2 (s) \rightleftharpoons I_2 (g) \) - **III)** \( H_2O (l) \rightleftharpoons H_2O (s) \) - **IV)** \( CO_2 (g) \rightleftharpoons CO_2 (aq) \) 2. **Analyze Each Reaction**: - **Equilibrium I**: The conversion of liquid water to water vapor is an endothermic process (absorbs heat). Therefore, an increase in temperature will shift the equilibrium to the right (towards the formation of water vapor). - **Equilibrium II**: The conversion of solid iodine to iodine vapor is also an endothermic process. Thus, an increase in temperature will shift the equilibrium to the right (towards the formation of iodine vapor). - **Equilibrium III**: The conversion of liquid water to solid ice is an exothermic process (releases heat). An increase in temperature will shift the equilibrium to the left (towards the formation of liquid water). - **Equilibrium IV**: The dissolution of carbon dioxide gas in water to form aqueous carbon dioxide is an exothermic process. An increase in temperature will shift the equilibrium to the left (favoring the formation of gaseous carbon dioxide). 3. **Conclusion**: - From the analysis, we see that Equilibria I and II shift to the right with an increase in temperature, while Equilibria III and IV shift to the left. Therefore, the correct answer is that the rise in temperature shifts the equilibrium towards the right in the case of **I and II**. ### Final Answer: The rise of temperature shifts equilibrium towards the right in the case of **I and II**.
Doubtnut Promotions Banner Mobile Dark
|

Similar Questions

Explore conceptually related problems

For the equilibrium H_(2) O (1) hArr H_(2) O (g) at 1 atm 298 K

C(s) +H_2O(g) hArr CO(g) +H_2(g) , DeltaH lt 0 the above equilibrium will proceed in forward direction when

State which one is homogeneous or heterogeneous? a. S_("Rhombus") hArr S_("Monoclinic") b. H_(2)O(l) hArr H_(2)O(v) c. H_(2)O(s) hArr H_(2)O(l) d. C_("Diamond") hArr C_("Amorphous") e. N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) f. CuSO_(4)(s)+3NH_(3)(g) hArr CuSO_(4).3 NH_(3)(l) g. CaCO(3)(s) hArr CaO(s)+CO_(2)(g)

Consider the following reactions: i. CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g), K_(1) ii. CH_(4)(g)+H_(2)O(g) hArr CO(g)+3H_(2)(g), K_(2) iii. CH_(4)(g)+2H_(2)O(g) hArr CO_(2)(g)+4H_(2)(g), K_(3) Which of the following is/are incorrect?

For the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) , the equilibrium can be shifted in favour of product by

For the reaction CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g) at a given temperature, the equilibrium amount of CO_(2)(g) can be increased by

{:(Lis - I, List -II),((A) H_(2) (g) + I_(2)(g) hArr 2HI(g),(P)"Unaffected by inert gas addition "),((B)CaCO_(3) (s) hArr CaO(s) + CO_(2) (g),(Q)" Forward shift by rise in pressure and backward shift by inert gas addition at constant pressure "),((C) N_(2) (g) + 3H_(2) + (g) hArr 2NH_(3) (g) , (R)" Unaffected by increase in pressure"),((D) PCl_(3) (g) hArr PCl_(3) (g) + Cl_(2) (g) ,(S)"ackward shift by rise in pressure and forward shift by inert gas addition at constant pressure "):}

{:(List - I , List - II),((A) N_(2) + O_(2) hArr 2NO,(P) (atm)^(-2)),((B) 2SO_(2)(g) + O_(2) hArr 2SO_(3)(g),(Q) "No unit"),((C) H_(2) + I_(2) hArr 2HI,(R)"No cffect of pressure"),((D) N_(2)(g) + 3H_(2) (g) hArr 2NH_(3)(g), (S) "High pressure favours forward reaction"):}

Consider the reaction: a. 6CO_(2)(g)+6H_(2)O(l)rarr C_(6)H_(12)O_(6)(aq)+6O_(2)(g) b. O_(3)(g)+H_(2)O_(2)(l) rarr H_(2)O(l)+2O_(2)(g) Why it is more appropriate to write these reaction as: a. 6CO_(2)(g)+12H_(2)O(l) rarr C_(6)H_(12)O_(6)(aq)+6H_(2)O(l)+6O_(2)(g) O_(3)(g)+H_(2)O_(2)(l) rarr H_(2)O(l)+O_(2)(g)+O_(2)(g) also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

If the equilibrium constant for N_(2) (g) + O_(2)(g) hArr 2NO(g) is K , the equilibrium " constant for " 1/2 N_(2) (g) +1/2 O_(2) (g) hArr NO (g) will be

Home
Profile