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Chemistry Question on Enthalpy change

At 25∘C25^{\circ} C, the enthalpy of the following processes are given: H2(g)+O2(g)→2OH(g)ΔH∘=78 kJ mol−1H _2( g )+ O _2( g ) \rightarrow 2 OH ( g ) \Delta H ^{\circ} =78 \,kJ \,mol ^{-1} H2(g)+1/2O2(g)→H2O(g)ΔH∘=−242 kJ mol−1H _2( g )+1 / 2 O _2( g ) \rightarrow H _2 O ( g ) \Delta H ^{\circ} =-242 \,kJ \,mol ^{-1} H2(g)→2H(g)ΔH∘=436 kJ mol−1H _2( g ) \rightarrow 2 H ( g ) \Delta H ^{\circ} =436\, kJ \,mol ^{-1} 1/2O2(g)→O(g)ΔH∘=249 kJ mol−11 / 2 O _2( g ) \rightarrow O ( g ) \Delta H ^{\circ} =249\, kJ\, mol ^{-1} What would be the value of XX for the following reaction?______ (Nearest integer) H2O(g)→H(g)+OH(g)ΔH∘=X kJ mol−1H _2 O ( g ) \rightarrow H ( g )+ OH ( g ) \Delta H ^{\circ}= X\, kJ\, mol -1

Answer

correct answer is 499.