At 25 o C , the enthalpy of the following processes are given: H 2 ( g ) + O 2 ( g ) 2 OH ( g ) H = 78 kJ mol - 1 H 2 ( g ) + 1 2 O 2 ( g ) H 2 O ( g ) H = - 242 kJ mol - 1 H 2 ( g ) 2 H ( g ) H ° = 436 kJ mol - 1

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
At $25^{o} C$ , the enthalpy of the following processes are given:

$H_{2} (g) + O_{2} (g) \to 2 OH (g) Δ H ° = {78 kJ mol}^{- 1}$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g) Δ H ° = - {242 kJ mol}^{- 1}$

$H_{2} (g) \to 2 H (g) Δ H ° = {436 kJ mol}^{- 1}$

$\frac{1}{2} O_{2} (g) \to O (g) Δ H ° = {249 kJ mol}^{- 1}$

What would be the value of X for the following reaction ________ (Nearest integer)

$H_{2} O (g) \to H (g) + OH (g) Δ H ° = X {kJ mol}^{- 1}$ [2023]

Ans.
(499)

$H_{2} O (g) \to H (g) + OH (g)$ $Δ H ° = {kJ mol}^{- 1}$

$2 H_{2} O \to 2 H_{2} + O_{2}$    $Δ H = + 242 \times 2 = 484$

$H_{2} + O_{2} \to 2 OH$ $Δ H = 78$

$2 H_{2} O \to H_{2} + 2 OH$    $Δ H = 562$

$H_{2} \to 2 H$ $Δ H = 436$

$2 H_{2} O \to 2 H + 2 OH$    $Δ H = 998$

$H_{2} O \to H + OH$    $Δ H = 499$

Want Us to Email you About Special Offers & Updates?