1. Tardigrade
  2. Question
  3. Chemistry
  4. At 25° C, the enthalpy of the following processes are given: H 2( g )+ O 2( g ) arrow 2 OH ( g ) Δ H ° =78 kJ mol -1 H 2( g )+1 / 2 O 2( g ) arrow H 2 O ( g ) Δ H ° =-242 kJ mol -1 H 2( g ) arrow 2 H ( g ) Δ H ° =436 kJ mol -1 1 / 2 O 2( g ) arrow 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 ) arrow H ( g )+ OH ( g ) Δ H °= X kJ mol -1

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Q. At $25^{\circ} C$, the enthalpy of the following processes are given:
$H _2( g )+ O _2( g ) \rightarrow 2 OH ( g ) \Delta H ^{\circ} =78 \,kJ \,mol ^{-1} $
$H _2( g )+1 / 2 O _2( g ) \rightarrow H _2 O ( g ) \Delta H ^{\circ} =-242 \,kJ \,mol ^{-1} $
$H _2( g ) \rightarrow 2 H ( g ) \Delta H ^{\circ} =436\, kJ \,mol ^{-1} $
$1 / 2 O _2( g ) \rightarrow O ( g ) \Delta H ^{\circ} =249\, kJ\, mol ^{-1}$
What would be the value of $X$ for the following reaction?______ (Nearest integer)
$H _2 O ( g ) \rightarrow H ( g )+ OH ( g ) \Delta H ^{\circ}= X\, kJ\, mol -1$

JEE MainJEE Main 2023Thermodynamics

Solution:

Correct answer is '499'

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