The reaction between hydrogen and fluorine is shown in the chemical equation below using displayed formulae:
H−H+F−F⟶2 H−F \text{H}-\text{H} + \text{F}-\text{F} \longrightarrow 2\ \text{H}-\text{F} H−H+F−F⟶2 H−FThe bond energies are given in the table below:
| Bond | Bond energy in kJ/mol\text{kJ/mol}kJ/mol |
|---|---|
| H−H\text{H}-\text{H}H−H | 436 |
| F−F\text{F}-\text{F}F−F | 158 |
| H−F\text{H}-\text{F}H−F | 562 |
Which expression shows how to calculate the overall energy change for this reaction?
436+158+562 kJ/mol436 + 158 + 562 \text{ kJ/mol}436+158+562 kJ/mol
436+158+(2×562) kJ/mol436 + 158 + (2 \times 562) \text{ kJ/mol}436+158+(2×562) kJ/mol
436+158−562 kJ/mol436 + 158 - 562 \text{ kJ/mol}436+158−562 kJ/mol
436+158−(2×562) kJ/mol436 + 158 - (2 \times 562) \text{ kJ/mol}436+158−(2×562) kJ/mol