The value of DHe for the reaction below is -790 kJ. The enthalp S is __________ kJ. 2S (s) + 3O2 (g) ¬ 2SO3 (g) ing the reaction of 0.95 g of S

To determine the enthalpy change (ΔH) for the reaction involving sulfur (S) and oxygen (O2) to form sulfur trioxide (SO3), we first need to understand the relationship between the amount of sulfur used and the given enthalpy change for the reaction. The reaction you've provided is:

Reaction Overview

2S (s) + 3O2 (g) → 2SO3 (g)

The enthalpy change (ΔH) for this reaction is given as -790 kJ for the formation of 2 moles of SO3 from 2 moles of sulfur and 3 moles of oxygen. This negative value indicates that the reaction is exothermic, meaning it releases energy.

Calculating the Enthalpy Change for 0.95 g of Sulfur

Next, we need to find out how much enthalpy change corresponds to the reaction of 0.95 g of sulfur. To do this, we will follow these steps:

• Step 1: Calculate the number of moles of sulfur.

The molar mass of sulfur (S) is approximately 32.07 g/mol. To find the number of moles in 0.95 g of sulfur, we use the formula:

Number of moles = mass (g) / molar mass (g/mol)

So, for 0.95 g of sulfur:

Number of moles = 0.95 g / 32.07 g/mol ≈ 0.0296 moles

• Step 2: Relate the moles of sulfur to the enthalpy change.

The balanced equation shows that 2 moles of sulfur release -790 kJ of energy. Therefore, the enthalpy change per mole of sulfur can be calculated as follows:

ΔH per mole of S = -790 kJ / 2 moles = -395 kJ/mole

• Step 3: Calculate the enthalpy change for 0.0296 moles of sulfur.

Now, we can find the enthalpy change for the 0.0296 moles of sulfur we calculated:

ΔH for 0.0296 moles = ΔH per mole × number of moles

ΔH for 0.0296 moles = -395 kJ/mole × 0.0296 moles ≈ -11.7 kJ

Final Result

Thus, the enthalpy change (ΔH) for the reaction of 0.95 g of sulfur is approximately -11.7 kJ. This means that when 0.95 g of sulfur reacts with oxygen to form sulfur trioxide, about 11.7 kJ of energy is released in the process.