Question 6.3: Manipulating Thermochemical Equations When 2 mol H2(g) and 1...

Manipulating Thermochemical Equations

When 2 mol H_2(g) and 1 mol O_2(g) react to give liquid water, 572 kJ of heat evolves.

2H_2(g)  +  O_2(g)  →  2H_2O(l );  ΔH  =  – 572  kJ

Write this equation for 1 mol of liquid water. Give the reverse equation, in which 1 mol of liquid water dissociates into hydrogen and oxygen.

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You multiply the coefficients and ΔH by \frac{1}{2} :

H _2(g) +  \frac{1}{2} O _2(g)  \longrightarrow  H _2 O (l) ;  \Delta H  =  -286  kJ

Reversing the equation, you get

H _2 O (l)  \longrightarrow  H _2(g)  +  \frac{1}{2} O _2(g) ;  \Delta H  =  +286  kJ

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