Question

Given the following equations and AH values, use Hess' law to determine the heat of reaction for the reaction: [App 4 marks]

2 OF2(g) + 2 S(s) → SO2(g) + SF4(g)

OF2(g) + H2O(l) → O2(g) + 2 HF(g) ΔH = – 276.6kJ
SF4(g) + 2 H2O(l) → 4 HF(g) + SO2(g) ΔH = -827.5kJ
S(s) + O2(g) → SO2(g) ΔH = – 296.9 kJ

Answer 1

To calculate the enthalpy change for the reaction 2 OF2(g) + 2 S(s) → SO2(g) + SF4(g) using Hess's Law, we manipulate and sum the given equations and their ΔH values, resulting in a total enthalpy change of 433.05 kJ.

Step-by-step explanation:

The question involves using Hess's Law to calculate the enthalpy change (ΔH) of a chemical reaction using given equations and their respective enthalpy changes. To find the enthalpy change for the reaction 2 OF2(g) + 2 S(s) → SO2(g) + SF4(g), we need to manipulate the given equations so that when added together, the result matches the target reaction.

First, we need to reverse the second equation and multiply it by ½ to align with the stoichiometry of SF4 in the target reaction:

• 1/2 (SF4(g) + 2 H2O(l) → 4 HF(g) + SO2(g), ΔH = +827.5/2 kJ)

Next, reverse the third equation:

• (SO2(g) → S(s) + O2(g), ΔH = +296.9 kJ)

We do not need to modify the first equation. Adding these three manipulated equations will result in the desired reaction, and the sum of their ΔH values will yield the ΔH for the target reaction:

• ΔH = (–276.6 kJ) + (+827.5/2 kJ) + (+296.9 kJ)

Now, we calculate the sum:

ΔH = -276.6 kJ + 413.75 kJ + 296.9 kJ = 433.05 kJ

So, the enthalpy change for the reaction 2 OF2(g) + 2 S(s) → SO2(g) + SF4(g) is 433.05 kJ.