The combustion of hexane takes place according to the equation $$C_6H_{14}(g) + \frac{19}{2} O_2(g) \rightarrow 6CO_2(g) + 7H_2O(g)$$ $$\Delta H = -4158 \text{ kJ mol}^{-1}$$ Consider the following reaction - VCE - SSCE Chemistry - Question 17 - 2016 - Paper 1

To find the enthalpy change for the new reaction, we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps.

We know from the original reaction:

$C_6H_{14} + \frac{19}{2} O_2 \rightarrow 6CO_2 + 7H_2O \quad \Delta H = -4158 \text{ kJ mol}^{-1}$

To find the reaction given:

$12CO_2 + 14H_2O \rightarrow 2C_6H_{14} + 19O$

We will reverse the original reaction and multiply it accordingly to match coefficients. Reversing the reaction changes the sign of \Delta H:

$6CO_2 + 7H_2O \rightarrow C_6H_{14} + \frac{19}{2} O_2 \quad \Delta H = +4158 \text{ kJ mol}^{-1}$

To match the new reaction, we will multiply by 2:

$12CO_2 + 14H_2O \rightarrow 2C_6H_{14} + 19O \quad \Delta H = 2 \times 4158 \text{ kJ mol}^{-1} = +8316 \text{ kJ mol}^{-1}$

Thus, the value of \Delta H for the given reaction is +8316 kJ mol^{-1}.