Chemistry · General chemistry I · Worked example

Reverse and scale a thermochemical equation

Liquid water forms from its elements with ΔH = −285.83 kJ: H₂(g) + ½O₂(g) → H₂O(l). Find ΔH for 2H₂O(l) → 2H₂(g) + O₂(g).

Δ H = - 2 Δ H_{f}^{}

Reverse the equation

Splitting water is its formation run backward, so the sign of ΔH flips: +285.83 kJ per mole of water.

H_{2} O (l) \to H_{2} (g) + \frac{1}{2} O_{2} (g)

Double it

The target has every coefficient doubled, so ΔH doubles too.

Δ H = 2 (+ 285.83 kJ) = + 571.66 kJ

\begin{array}{c} Δ H = 2 (+ 285.83 kJ) \\ = + 571.66 kJ \end{array}

Interpret the sign

The reaction is endothermic: splitting 2 mol of liquid water takes in 571.66 kJ, exactly what forming it gives out.

Result

ΔH = +571.66 kJ for 2H₂O(l) → 2H₂(g) + O₂(g).

Your turn

How much heat does burning 1.00 g of hydrogen gas to liquid water release?

Show the answer and explanation

142 kJ is released.

1.00 g of H₂ is 1.00 ÷ 2.016 = 0.496 mol, and each mole releases 285.83 kJ as it forms liquid water: 0.496 × 285.83 = 142 kJ.

1.00 g H_{2} \times \frac{1 mol H_{2}}{2.016 g H_{2}} \times \frac{285.83 kJ}{1 mol H_{2}} = 142 kJ

\begin{array}{c} 1.00 g H_{2} \times \frac{1 mol H_{2}}{2.016 g H_{2}} \\ \times \frac{285.83 kJ}{1 mol H_{2}} = 142 kJ \end{array}

Keep exploring

In Hess law & thermodynamics, change the multiplier to −1 and the target to H₂O(l) → H₂(g) + ½O₂(g). ΔH halves to +285.83 kJ, for splitting one mole of water.

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OpenStax Chemistry 2e — Enthalpy

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